💡Okay, so I have this geometry problem here about triangle ABC. The problem states that AF is twice BF, CE is three times AE, and CD is four times BD. Then, we're supposed to connect CF and see where it intersects DE. Hmm, okay, let me try to visualize this.First, let me draw triangle ABC. Let me label the vertices A, B, and C. Now, AF is twice BF, so point F must be on side AB. Since AF is twice BF, that means F divides AB into segments AF and BF with AF being twice as long as BF. So, if I let BF be x, then AF would be 2x, making the entire length of AB equal to 3x.Similarly, CE is three times AE. So, point E must be on side AC. If I let AE be y, then CE would be 3y, making the entire length of AC equal to 4y.Next, CD is four times BD. So, point D must be on side BC. If I let BD be z, then CD would be 4z, making the entire length of BC equal to 5z.Now, we need to connect CF and see where it intersects DE. So, CF is a line from C to F on AB, and DE is a line from D on BC to E on AC. Their intersection point is what we're interested in.I think I need to use coordinate geometry to solve this. Let me assign coordinates to the triangle. Let me place point A at (0, 0), point B at (3, 0), and point C at (0, 4). This way, AB is along the x-axis from (0,0) to (3,0), AC is along the y-axis from (0,0) to (0,4), and BC connects (3,0) to (0,4).Given this coordinate system, let me find the coordinates of points F, E, and D.Starting with point F on AB. Since AF is twice BF, and AB is from (0,0) to (3,0), the total length is 3 units. So, AF = 2 units, BF = 1 unit. Therefore, point F is located at (2, 0).Next, point E on AC. CE is three times AE, so AC is 4 units. Therefore, AE = 1 unit, CE = 3 units. So, point E is located at (0, 1).Now, point D on BC. CD is four times BD, so BC is 5 units. Therefore, BD = 1 unit, CD = 4 units. To find the coordinates of D, I need to find the point that divides BC in the ratio BD:DC = 1:4.The coordinates of B are (3, 0) and C are (0, 4). Using the section formula, the coordinates of D can be calculated as:D = [(4*3 + 1*0)/(1+4), (4*0 + 1*4)/(1+4)] = [(12 + 0)/5, (0 + 4)/5] = (12/5, 4/5).So, D is at (12/5, 4/5).Now, I need to find the equations of lines CF and DE and then find their intersection point.First, let's find the equation of line CF. Point C is at (0, 4) and point F is at (2, 0). The slope of CF is (0 - 4)/(2 - 0) = (-4)/2 = -2. So, the equation of CF is y - 4 = -2(x - 0), which simplifies to y = -2x + 4.Next, let's find the equation of line DE. Point D is at (12/5, 4/5) and point E is at (0, 1). The slope of DE is (1 - 4/5)/(0 - 12/5) = (1/5)/(-12/5) = -1/12. So, the equation of DE is y - 1 = (-1/12)(x - 0), which simplifies to y = (-1/12)x + 1.Now, to find the intersection point P of CF and DE, we can set the equations equal to each other:-2x + 4 = (-1/12)x + 1.Let's solve for x:-2x + 4 = (-1/12)x + 1 Multiply both sides by 12 to eliminate the fraction:-24x + 48 = -x + 12 Bring all terms to one side:-24x + 48 + x - 12 = 0 -23x + 36 = 0 -23x = -36 x = (-36)/(-23) = 36/23.Now, substitute x = 36/23 into one of the equations to find y. Let's use the equation of CF: y = -2x + 4.y = -2*(36/23) + 4 = -72/23 + 92/23 = (92 - 72)/23 = 20/23.So, the intersection point P is at (36/23, 20/23).Now, the problem asks for the ratio in which P divides DE. So, we need to find the ratio EP:PD.Point E is at (0,1), point P is at (36/23, 20/23), and point D is at (12/5, 4/5).Let me calculate the distances EP and PD.First, let's find the distance EP. Using the distance formula:EP = sqrt[(36/23 - 0)^2 + (20/23 - 1)^2] = sqrt[(36/23)^2 + (-3/23)^2] = sqrt[(1296/529) + (9/529)] = sqrt[1305/529] = sqrt[1305]/23.Similarly, PD = sqrt[(12/5 - 36/23)^2 + (4/5 - 20/23)^2].First, let's compute the differences:12/5 - 36/23 = (12*23 - 36*5)/(5*23) = (276 - 180)/115 = 96/115.4/5 - 20/23 = (4*23 - 20*5)/(5*23) = (92 - 100)/115 = (-8)/115.So, PD = sqrt[(96/115)^2 + (-8/115)^2] = sqrt[(9216/13225) + (64/13225)] = sqrt[9280/13225] = sqrt[9280]/115.Simplify sqrt[9280]: 9280 = 16 * 580 = 16 * 4 * 145 = 64 * 145. So, sqrt[9280] = 8*sqrt[145].Thus, PD = 8*sqrt[145]/115.Similarly, EP was sqrt[1305]/23. Let's see if sqrt[1305] can be simplified:1305 = 5 * 261 = 5 * 3 * 87 = 5 * 3 * 3 * 29 = 9 * 5 * 29. So, sqrt[1305] = 3*sqrt[145].Therefore, EP = 3*sqrt[145]/23.Now, let's find the ratio EP:PD.EP = 3*sqrt[145]/23 PD = 8*sqrt[145]/115So, EP:PD = (3*sqrt[145]/23) : (8*sqrt[145]/115)We can simplify this ratio by dividing both parts by sqrt[145]:= (3/23) : (8/115)To divide these fractions, we can multiply the first by the reciprocal of the second:= (3/23) * (115/8) = (3 * 115) / (23 * 8) = (3 * 5) / 8 = 15/8.So, the ratio EP:PD is 15:8.Wait, that seems a bit high. Let me double-check my calculations.First, when calculating PD, I had:12/5 - 36/23 = (276 - 180)/115 = 96/115. 4/5 - 20/23 = (92 - 100)/115 = -8/115.So, PD = sqrt[(96/115)^2 + (-8/115)^2] = sqrt[(9216 + 64)/13225] = sqrt[9280/13225] = sqrt[9280]/115.Then, sqrt[9280] = sqrt[64*145] = 8*sqrt[145]. So, PD = 8*sqrt[145]/115.Similarly, EP was sqrt[(36/23)^2 + (-3/23)^2] = sqrt[(1296 + 9)/529] = sqrt[1305]/23 = 3*sqrt[145]/23.So, EP:PD = (3*sqrt[145]/23) : (8*sqrt[145]/115) = (3/23) : (8/115) = (3/23) * (115/8) = (3*5)/8 = 15/8.Yes, that seems correct. So, the ratio EP:PD is 15:8.Alternatively, maybe there's a way to solve this using mass point geometry or area ratios instead of coordinate geometry, which might be quicker.Let me try using area ratios.In triangle ABC, AF:FB = 2:1, so the area of triangle AFC is twice that of BFC.Similarly, CE:EA = 3:1, so the area of triangle CEF is three times that of AEF.And CD:DB = 4:1, so the area of triangle CFD is four times that of BFD.Let me denote the area of triangle ABC as S.Then, area of AFC = (2/3)S, area of BFC = (1/3)S.Area of CEF = (3/4)*area of AFC = (3/4)*(2/3)S = (1/2)S.Area of CFD = (4/5)*area of BFC = (4/5)*(1/3)S = (4/15)S.Now, the ratio of areas of triangles EFC and FDC is (1/2)S : (4/15)S = (1/2)/(4/15) = (15/8).Since these triangles share the same base FC, the ratio of their areas is equal to the ratio of their heights from E and D to FC.Therefore, the ratio EP:PD is 15:8.Yes, that confirms the earlier result.So, the ratio in which P divides DE is 15:8.

💡Okay, so I have this problem where there's a 1-meter long wire, and somewhere along it, there's a break that stops it from conducting electricity. The break isn't visible, so I can't just look for it. The teacher wants me to figure out a quick way to locate the break without checking every single segment one by one, which would be really time-consuming. The goal is to narrow down the location of the break to about 3-4 centimeters. I need to figure out how many times I need to check to achieve that.First, I need to understand the problem better. A 1-meter wire is 100 centimeters long. There's a break somewhere in this wire, and I need to find where it is. Since it's a break in the wire, it means that at some point, the wire stops conducting electricity. So, if I can test for continuity at different points along the wire, I can determine where the break is.The initial thought is to check the wire segment by segment. That would mean dividing the wire into small segments, say 1 centimeter each, and checking each one until I find the break. But the problem says that this would be quite troublesome, so I need a more efficient method.I remember hearing about something called "binary search" in computer science, which is a way to find an item in a sorted list by repeatedly dividing the search interval in half. Maybe I can apply that idea here. Instead of checking every single segment, I can check the midpoint of the wire and then narrow down the search area based on the result.Let me think about how that would work. If I start with the entire 100-centimeter wire, I can check at the 50-centimeter mark. If there's continuity from one end to the 50-centimeter point, then the break must be somewhere between 50 and 100 centimeters. If there's no continuity, then the break is somewhere between 0 and 50 centimeters. Either way, I've reduced the search area by half with just one check.If I continue this process, each time checking the midpoint of the current search area, I can keep narrowing down the location of the break. This seems like a much more efficient method than checking every single segment.Now, the question is, how many times do I need to check to narrow down the location to about 3-4 centimeters? Let's break this down step by step.1. **First Check**: Start with the entire 100 cm wire. Check at the 50 cm mark. - If there's continuity, the break is between 50 cm and 100 cm. - If there's no continuity, the break is between 0 cm and 50 cm. - After this check, the search area is reduced to 50 cm.2. **Second Check**: Take the 50 cm segment identified in the first check and check at the 25 cm mark within that segment. - If there's continuity, the break is between 25 cm and 50 cm. - If there's no continuity, the break is between 0 cm and 25 cm. - Now, the search area is 25 cm.3. **Third Check**: Take the 25 cm segment and check at the 12.5 cm mark. - If there's continuity, the break is between 12.5 cm and 25 cm. - If there's no continuity, the break is between 0 cm and 12.5 cm. - The search area is now 12.5 cm.4. **Fourth Check**: Take the 12.5 cm segment and check at the 6.25 cm mark. - If there's continuity, the break is between 6.25 cm and 12.5 cm. - If there's no continuity, the break is between 0 cm and 6.25 cm. - The search area is now 6.25 cm.5. **Fifth Check**: Take the 6.25 cm segment and check at the 3.125 cm mark. - If there's continuity, the break is between 3.125 cm and 6.25 cm. - If there's no continuity, the break is between 0 cm and 3.125 cm. - The search area is now approximately 3.125 cm.At this point, after five checks, the search area is about 3.125 cm, which is within the desired range of 3-4 centimeters. Therefore, it seems like five checks are sufficient to narrow down the location of the break to about 3-4 centimeters.But wait, let me double-check this. Each check effectively halves the search area. Starting from 100 cm, after one check, it's 50 cm; after two checks, 25 cm; three checks, 12.5 cm; four checks, 6.25 cm; and five checks, 3.125 cm. Yes, that seems correct.Is there a way to do it in fewer checks? Let's see. If I try to do it in four checks, the search area would be 6.25 cm, which is larger than the desired 3-4 cm. So, five checks are necessary to get within the desired range.Another thought: what if the wire isn't perfectly divisible by halves each time? For example, if I have an odd number of centimeters, like 101 cm, but in this case, it's exactly 100 cm, so it's fine. Also, in practice, measuring exactly at the midpoint might be a bit tricky, but for the sake of this problem, we can assume that we can measure precisely.Also, considering that each check involves testing for continuity, which is a binary result (either there is continuity or there isn't), this aligns perfectly with the binary search method, where each step gives us a yes/no answer that halves the search space.So, to summarize, by using a binary search approach, I can efficiently narrow down the location of the break in the wire. Each check reduces the search area by half, and after five checks, the search area is approximately 3.125 cm, which meets the requirement of narrowing it down to about 3-4 centimeters.I think this makes sense. It's a logical approach, and it's much more efficient than checking each segment individually. Plus, it's a practical application of the binary search algorithm, which is a fundamental concept in computer science and problem-solving.One last thing to consider: what if the break isn't exactly at a midpoint? Well, the binary search method still works because it's about narrowing down the possible location, not pinpointing it exactly. So, even if the break isn't exactly at a midpoint, the method still effectively reduces the search area each time, bringing us closer to the exact location.In conclusion, using the binary search method, I can locate the break in the wire with five checks, narrowing down the location to approximately 3-4 centimeters.

💡Okay, so I have this problem here where each data point in a set is multiplied by 2 and then reduced by 3, resulting in a new set of data. I'm told that the average of this new data is 7 and the variance is 4. I need to find the average and variance of the original data set.Alright, let's break this down. First, let's think about what happens when we transform each data point in a set. If I have an original data set, say, {x₁, x₂, ..., xₙ}, and I apply a linear transformation to each data point, like multiplying by 2 and then subtracting 3, the new data set becomes {2x₁ - 3, 2x₂ - 3, ..., 2xₙ - 3}.Now, I remember that when you apply a linear transformation to each data point, it affects both the average and the variance of the data set. Specifically, if you multiply each data point by a constant, the average gets multiplied by that same constant, and the variance gets multiplied by the square of that constant. Then, if you add or subtract a constant from each data point, it shifts the average but doesn't affect the variance because variance measures the spread around the average, not the location of the average.So, in this case, each data point is multiplied by 2 and then reduced by 3. That means the average of the new data set should be 2 times the original average minus 3. Similarly, the variance of the new data set should be 2 squared times the original variance, which is 4 times the original variance.Given that the average of the new data set is 7, I can set up an equation to solve for the original average. Let me denote the original average as μ. Then, according to the transformation, 2μ - 3 = 7. Solving for μ, I can add 3 to both sides to get 2μ = 10, and then divide both sides by 2 to find μ = 5. So, the original average should be 5.Now, moving on to the variance. The variance of the new data set is given as 4. Since variance scales with the square of the multiplier, and we multiplied each data point by 2, the variance of the new data set is 4 times the original variance. Let me denote the original variance as σ². Then, according to the transformation, 4σ² = 4. To find σ², I can divide both sides by 4, which gives σ² = 1. So, the original variance should be 1.Wait, let me double-check that. If I have a data set and I multiply each point by 2, the spread of the data doubles, so the variance, which is the square of the standard deviation, should increase by the square of the multiplier. That makes sense because variance is in squared units. So, multiplying by 2 increases the spread, hence the variance, by a factor of 4. Then, subtracting 3 just shifts all the data points but doesn't change how spread out they are, so the variance remains the same after the shift.Therefore, my calculations seem correct. The original average is 5, and the original variance is 1.Just to be thorough, let me consider an example. Suppose the original data set has an average of 5 and a variance of 1. If I multiply each data point by 2, the new average becomes 10, and the new variance becomes 4. Then, if I subtract 3 from each data point, the new average becomes 7, and the variance remains 4. That matches the given conditions in the problem. So, my answer seems consistent.I think I've got it. The original average is 5, and the original variance is 1.

💡Okay, so I've got this problem here about mappings between two sets, x and y. Let me try to understand what it's asking.First, the sets are defined as x = {-1, 0, 1} and y = {-2, -1, 0, 1, 2}. I need to find how many mappings f from x to y satisfy the condition that for every element x in the set, both x and x + f(x) are even. Hmm, okay, so for each x in {-1, 0, 1}, both x and x + f(x) must be even numbers. Wait, does that mean that x and x + f(x) have to be both even or both odd? Because the problem says "both x and x + f(x) are even." So, does that mean they have to be even together, or does it mean that each of them individually is even? Let me read it again: "both x and x + f(x) are even." So, I think it means that each of them is even. So, for each x, x is even and x + f(x) is even. But hold on, x is given as {-1, 0, 1}. So, let's check the parity of each x:- x = -1: That's odd.- x = 0: That's even.- x = 1: That's odd.So, for x = -1 and x = 1, which are odd, the problem says that both x and x + f(x) are even. But x is odd, so x + f(x) must also be odd? Wait, no, because if x is odd, and x + f(x) is even, then f(x) must be odd. Because odd + odd = even. Similarly, if x is even, then x + f(x) must be even, so f(x) must be even.Wait, so maybe I misread the problem. It says "both x and x + f(x) are even." So, for each x, both x and x + f(x) must be even. But x is given as {-1, 0, 1}, so x can be odd or even. So, for x = -1 and x = 1, which are odd, x + f(x) must be even. So, since x is odd, f(x) must be odd to make x + f(x) even. For x = 0, which is even, x + f(x) must be even, so f(x) must be even.Wait, so let me rephrase:For each x in {-1, 0, 1}:- If x is even (which is only x=0), then x + f(x) must be even. Since x is even, f(x) must also be even.- If x is odd (which are x=-1 and x=1), then x + f(x) must be even. Since x is odd, f(x) must be odd.So, f(x) must be even when x is even, and odd when x is odd.So, for x=0, f(0) must be even. For x=-1 and x=1, f(-1) and f(1) must be odd.Now, let's look at the set y = {-2, -1, 0, 1, 2}. Let's list the even and odd numbers in y:Even numbers in y: -2, 0, 2Odd numbers in y: -1, 1So, for x=0, f(0) can be any even number in y, which are -2, 0, 2. So, 3 choices.For x=-1 and x=1, f(-1) and f(1) must be odd numbers in y, which are -1 and 1. So, each has 2 choices.Therefore, the total number of mappings f is the product of the number of choices for each x.So, total mappings = number of choices for f(0) * number of choices for f(-1) * number of choices for f(1) = 3 * 2 * 2 = 12.Wait, but let me double-check. Is there any restriction I'm missing? The problem says "mappings from x to y," so each element in x is mapped to an element in y. So, for each x, f(x) is chosen independently, right? So, yes, 3 choices for f(0), 2 for f(-1), and 2 for f(1), so 3*2*2=12.But let me make sure I didn't misinterpret the condition. The problem says "both x and x + f(x) are even." So, for x=0, which is even, x + f(x) must be even, so f(x) must be even. For x=-1 and x=1, which are odd, x + f(x) must be even, so f(x) must be odd. That seems correct.So, the total number of such mappings is 12. Looking at the options, C is 12, so I think that's the answer.Wait, but let me think again. Is there a possibility that x + f(x) being even could have another interpretation? Like, maybe x and x + f(x) are both even or both odd? But the problem says "both x and x + f(x) are even," so I think it's specifically that both are even, not that they have the same parity. So, if x is odd, then x + f(x) must be even, which requires f(x) to be odd. If x is even, then x + f(x) must be even, which requires f(x) to be even.Yes, that seems correct. So, the answer should be 12, which is option C.**Final Answer**boxed{C}

💡Okay, so I have this geometry problem here about a parallelogram ABCD. Let me try to visualize it first. In a parallelogram, opposite sides are equal and parallel, and opposite angles are equal. So, angle DAB is equal to angle BCD, and angle ABC is equal to angle CDA. Also, the sum of consecutive angles is 180 degrees, right? So, if angle DAB is, say, θ, then angle ABC would be 180 - θ.Now, there's a point E on BC, between B and C. So, E is somewhere along the side BC. The problem states that triangles DEC, BED, and BAD are all isosceles. Hmm, okay. Let me break this down.First, triangle DEC is isosceles. That means two of its sides are equal. So, either DE = EC, DE = DC, or EC = DC. But since ABCD is a parallelogram, DC is equal to AB, which is a side of the parallelogram. So, EC can't be equal to DC unless E is at point C, but E is between B and C, so EC is less than DC. Therefore, EC can't be equal to DC. So, either DE = EC or DE = DC.Similarly, triangle BED is isosceles. So, either BE = ED, BE = BD, or ED = BD. Again, BD is a diagonal of the parallelogram. In a parallelogram, the diagonals bisect each other but aren't necessarily equal unless it's a rectangle. So, unless ABCD is a rectangle, BD isn't equal to AC. But we don't know that yet. So, BE = ED or BE = BD or ED = BD.Lastly, triangle BAD is isosceles. So, either BA = AD, BA = BD, or AD = BD. Since BA and AD are sides of the parallelogram, BA = CD and AD = BC. So, if BA = AD, then AB = AD, meaning the parallelogram is a rhombus. Alternatively, if BA = BD or AD = BD, that would mean the sides are equal to the diagonal, which is a bit more complex.Let me try to approach this step by step.First, let's consider triangle BAD being isosceles. So, either BA = AD, BA = BD, or AD = BD.Case 1: BA = AD. If BA = AD, then the parallelogram is a rhombus because all sides are equal. In a rhombus, all sides are equal, and the diagonals bisect the angles. So, if ABCD is a rhombus, then the diagonals AC and BD intersect at right angles and bisect each other.Case 2: BA = BD. So, the side BA is equal to the diagonal BD. That would impose some specific angle conditions. Similarly, Case 3: AD = BD. So, side AD equals diagonal BD.But let's see if we can figure out which case it is based on the other triangles being isosceles.Let's move to triangle DEC. It's isosceles, so either DE = EC or DE = DC.If DE = EC, then E is the midpoint of DC. But wait, E is on BC, not DC. Hmm, that might not make sense. Wait, no, in a parallelogram, DC is opposite to AB, so DC is equal to AB. But E is on BC, so EC is a segment from E to C on BC. So, if DE = EC, then E is somewhere on BC such that the length from E to C is equal to the length from D to E.Alternatively, if DE = DC, then DE is equal to DC, which is equal to AB. So, DE would be equal to AB.Similarly, for triangle BED being isosceles, either BE = ED, BE = BD, or ED = BD.Let me try to sketch this mentally. Let me assume that triangle BAD is isosceles with BA = AD. So, ABCD is a rhombus. Then, all sides are equal, and the diagonals bisect each other at right angles.In that case, point E is on BC. Let's see what happens with triangles DEC and BED.If ABCD is a rhombus, then BC is equal to AB, which is equal to AD. So, BC is equal to AB. Now, E is on BC. Let's say E is somewhere between B and C.Now, triangle DEC: if it's isosceles, then either DE = EC or DE = DC. Since DC is equal to AB, which is equal to BC, so DC is equal to BC. So, if DE = DC, then DE = BC. But E is on BC, so DE would have to be equal to BC. That might be possible if E is at a specific point.Alternatively, DE = EC. So, E is such that DE = EC. That would mean that E is located such that the distance from E to C is equal to the distance from D to E. Hmm, that might place E somewhere specific on BC.Similarly, triangle BED is isosceles. So, either BE = ED, BE = BD, or ED = BD.If BE = ED, then E is such that the distance from B to E is equal to the distance from E to D. That would place E somewhere on BC such that BE = ED.Alternatively, if BE = BD, then E would have to be such that BE equals the diagonal BD. But in a rhombus, the diagonals are longer than the sides, so BE can't be equal to BD unless E is beyond point C, which it's not. So, that's not possible.Similarly, ED = BD would mean that the distance from E to D is equal to the diagonal BD, which is longer than the sides, so again, E would have to be beyond C, which it's not. So, in this case, the only possibility is BE = ED.So, in the rhombus case, we have triangle BAD is isosceles with BA = AD, triangle DEC is isosceles with DE = EC, and triangle BED is isosceles with BE = ED.Wait, but if E is such that BE = ED and DE = EC, then BE = ED = EC. So, E is equidistant from B and C, and also from D and C. Hmm, that might only happen if E is the midpoint of BC. Because if E is the midpoint, then BE = EC, but we also have DE = EC, so DE = BE. So, E is the midpoint.Wait, but in a rhombus, the diagonals bisect each other at right angles. So, the midpoint of BC would be the point where the diagonal intersects BC. But in a rhombus, the diagonals bisect each other, but they don't necessarily bisect the sides unless it's a square.Wait, no, in a rhombus, the diagonals bisect the angles, but they don't necessarily bisect the sides. So, the midpoint of BC is not necessarily where the diagonal intersects BC. Hmm, so maybe E is the midpoint of BC, but in a rhombus, that's not where the diagonal intersects.Wait, let me think. In a rhombus, the diagonals bisect each other, so they intersect at the midpoint of both diagonals. So, the midpoint of diagonal AC is the same as the midpoint of diagonal BD. But the midpoint of BC is a different point.So, if E is the midpoint of BC, then BE = EC. But in the rhombus, the diagonals don't pass through the midpoints of the sides unless it's a square.So, maybe E is not the midpoint of BC, but somewhere else.Alternatively, maybe the rhombus is a square. If ABCD is a square, then all sides are equal, and all angles are 90 degrees. Then, E is on BC. Let's see.In a square, triangle DEC being isosceles: if E is on BC, then DE and EC would have to be equal. So, DE = EC. Similarly, triangle BED being isosceles: BE = ED or BE = BD or ED = BD. But BD is the diagonal, which is longer than the sides, so BE can't equal BD unless E is beyond C, which it's not. So, BE = ED.So, in a square, if E is such that BE = ED and DE = EC, then E is the midpoint of BC. Because if BE = ED and DE = EC, then BE = EC, so E is the midpoint.Wait, but in a square, if E is the midpoint of BC, then DE would be equal to EC. Let me check.In a square, BC is a side of length, say, 'a'. So, BE = EC = a/2. Now, DE is the distance from D to E. Let's place the square on a coordinate system to make it easier.Let me assign coordinates: Let’s say A is at (0,0), B at (a,0), C at (a,a), and D at (0,a). Then, E is on BC, which goes from (a,0) to (a,a). So, E is at (a, e), where e is between 0 and a.Now, DE is the distance from D(0,a) to E(a,e), which is sqrt((a-0)^2 + (e - a)^2) = sqrt(a² + (e - a)²).EC is the distance from E(a,e) to C(a,a), which is sqrt((a - a)^2 + (a - e)^2) = sqrt(0 + (a - e)²) = |a - e|.So, if DE = EC, then sqrt(a² + (e - a)²) = |a - e|.Squaring both sides: a² + (e - a)² = (a - e)².But (e - a)² is the same as (a - e)², so we have a² + (a - e)² = (a - e)².Subtracting (a - e)² from both sides: a² = 0, which implies a = 0, but that's not possible because a is the side length of the square.So, DE cannot be equal to EC in a square unless a = 0, which is impossible. Therefore, in a square, triangle DEC cannot be isosceles with DE = EC.Hmm, that's a problem. So, maybe the assumption that ABCD is a rhombus is incorrect.Wait, but triangle BAD is isosceles. If BA = AD, then it's a rhombus. If BA = BD or AD = BD, then it's a different case.Let me consider the case where triangle BAD is isosceles with BA = BD. So, BA = BD.In a parallelogram, BA is a side, and BD is a diagonal. So, if BA = BD, then the diagonal BD is equal in length to the side BA.In a parallelogram, the length of the diagonal can be found using the formula: BD² = AB² + AD² - 2AB*AD*cos(angle BAD).Wait, no, actually, in a parallelogram, the sum of the squares of the diagonals equals twice the sum of the squares of the sides: AC² + BD² = 2(AB² + AD²).But if BA = BD, then BD = AB. So, let's denote AB = x, then BD = x.So, AC² + x² = 2(x² + AD²).But in a parallelogram, AD = BC, and AB = CD.Wait, but we don't know AD yet. Let me denote AD = y.So, AC² + x² = 2(x² + y²).But we also know that in triangle ABD, which is triangle BAD, if BA = BD, then triangle BAD is isosceles with BA = BD.Wait, but in triangle BAD, BA and BD are two sides, and AD is the base.So, in triangle BAD, sides BA and BD are equal, so angles opposite them are equal. So, angle BAD is equal to angle BDA.But in a parallelogram, angle BAD is equal to angle BCD, and angle ABC is equal to angle CDA.Hmm, this is getting a bit complicated. Maybe I should try to use coordinates again.Let me assign coordinates to the parallelogram. Let’s place point A at (0,0). Since it's a parallelogram, let me denote point B as (a,0), point D as (b,c), and point C as (a + b, c).Now, E is a point on BC. Let's parameterize E. Since BC goes from (a,0) to (a + b, c), we can write E as (a + tb, tc), where t is between 0 and 1.Now, let's consider triangle DEC. Points D(b,c), E(a + tb, tc), and C(a + b, c).So, the distances:DE: distance between D(b,c) and E(a + tb, tc) is sqrt[(a + tb - b)^2 + (tc - c)^2] = sqrt[(a + b(t - 1))^2 + (c(t - 1))^2].EC: distance between E(a + tb, tc) and C(a + b, c) is sqrt[(a + b - a - tb)^2 + (c - tc)^2] = sqrt[(b(1 - t))^2 + (c(1 - t))^2] = (1 - t)sqrt(b² + c²).DC: distance between D(b,c) and C(a + b, c) is sqrt[(a + b - b)^2 + (c - c)^2] = sqrt[a² + 0] = a.So, triangle DEC is isosceles. So, either DE = EC, DE = DC, or EC = DC.Case 1: DE = EC.So, sqrt[(a + b(t - 1))² + (c(t - 1))²] = (1 - t)sqrt(b² + c²).Squaring both sides:(a + b(t - 1))² + (c(t - 1))² = (1 - t)²(b² + c²).Expanding the left side:[a + b(t - 1)]² + [c(t - 1)]² = [a² + 2ab(t - 1) + b²(t - 1)²] + [c²(t - 1)²] = a² + 2ab(t - 1) + (b² + c²)(t - 1)².The right side is (1 - t)²(b² + c²) = (t - 1)²(b² + c²).So, setting them equal:a² + 2ab(t - 1) + (b² + c²)(t - 1)² = (t - 1)²(b² + c²).Subtracting the right side from both sides:a² + 2ab(t - 1) = 0.So, a² + 2ab(t - 1) = 0.Solving for t:2ab(t - 1) = -a²t - 1 = -a² / (2ab) = -a / (2b)t = 1 - a/(2b)So, t = (2b - a)/(2b).Since t must be between 0 and 1, we have:(2b - a)/(2b) ≥ 0 and ≤ 1.So, 2b - a ≥ 0 ⇒ a ≤ 2b.And (2b - a)/(2b) ≤ 1 ⇒ 2b - a ≤ 2b ⇒ -a ≤ 0 ⇒ a ≥ 0, which is always true since a is a length.So, t = (2b - a)/(2b).Now, let's consider triangle BED. Points B(a,0), E(a + tb, tc), D(b,c).So, distances:BE: distance between B(a,0) and E(a + tb, tc) is sqrt[(tb)^2 + (tc)^2] = t sqrt(b² + c²).ED: distance between E(a + tb, tc) and D(b,c) is sqrt[(a + tb - b)^2 + (tc - c)^2] = sqrt[(a + b(t - 1))^2 + (c(t - 1))^2], which we already calculated as DE.BD: distance between B(a,0) and D(b,c) is sqrt[(b - a)^2 + c²].So, triangle BED is isosceles. So, either BE = ED, BE = BD, or ED = BD.Case 1a: BE = ED.So, t sqrt(b² + c²) = sqrt[(a + b(t - 1))² + (c(t - 1))²].But from earlier, we have DE = EC, which led us to t = (2b - a)/(2b).So, substituting t into BE = ED:t sqrt(b² + c²) = sqrt[(a + b(t - 1))² + (c(t - 1))²].But from the earlier equation, we know that DE = EC, which led to a² + 2ab(t - 1) = 0, so a² = -2ab(t - 1).So, a² = -2ab(t - 1) ⇒ a = -2b(t - 1) ⇒ a = 2b(1 - t).So, t = 1 - a/(2b).So, substituting t into BE = ED:t sqrt(b² + c²) = sqrt[(a + b(t - 1))² + (c(t - 1))²].But from a = 2b(1 - t), we can express a in terms of t and b.Let me substitute a = 2b(1 - t) into the equation.So, t sqrt(b² + c²) = sqrt[(2b(1 - t) + b(t - 1))² + (c(t - 1))²].Simplify the expression inside the sqrt:2b(1 - t) + b(t - 1) = 2b - 2bt + bt - b = (2b - b) + (-2bt + bt) = b - bt = b(1 - t).Similarly, c(t - 1) = -c(1 - t).So, the expression becomes sqrt[(b(1 - t))² + (-c(1 - t))²] = sqrt[(b² + c²)(1 - t)²] = (1 - t)sqrt(b² + c²).So, the equation becomes:t sqrt(b² + c²) = (1 - t)sqrt(b² + c²).Divide both sides by sqrt(b² + c²):t = 1 - t ⇒ 2t = 1 ⇒ t = 1/2.So, t = 1/2.But from earlier, t = (2b - a)/(2b).So, 1/2 = (2b - a)/(2b) ⇒ (2b - a) = b ⇒ 2b - a = b ⇒ a = b.So, a = b.So, in this case, a = b.So, the coordinates become:A(0,0), B(a,0), D(a,c), C(2a,c).E is at (a + tb, tc) = (a + (1/2)a, (1/2)c) = (3a/2, c/2).Wait, but in a parallelogram, point C should be at (a + b, c). Since a = b, C is at (2a, c).So, E is at (3a/2, c/2).Now, let's check triangle BAD. Points B(a,0), A(0,0), D(a,c).So, BA is from (0,0) to (a,0), length a.AD is from (0,0) to (a,c), length sqrt(a² + c²).BD is from (a,0) to (a,c), length c.So, triangle BAD has sides BA = a, AD = sqrt(a² + c²), BD = c.We were told that triangle BAD is isosceles. So, either BA = AD, BA = BD, or AD = BD.Case 1: BA = AD ⇒ a = sqrt(a² + c²). But sqrt(a² + c²) > a unless c = 0, which would make it a degenerate parallelogram (a line). So, this is not possible.Case 2: BA = BD ⇒ a = c.Case 3: AD = BD ⇒ sqrt(a² + c²) = c ⇒ a² + c² = c² ⇒ a² = 0 ⇒ a = 0, which is not possible.So, the only possibility is BA = BD ⇒ a = c.So, a = c.So, in this case, a = b = c.So, the coordinates become:A(0,0), B(a,0), D(a,a), C(2a,a).E is at (3a/2, a/2).Now, let's check the angles.In parallelogram ABCD, angle DAB is the angle at A between sides AB and AD.Vector AB is (a,0), vector AD is (a,a).The angle between AB and AD can be found using the dot product:cos(theta) = (AB · AD) / (|AB||AD|) = (a*a + 0*a) / (a * sqrt(a² + a²)) = a² / (a * a√2) = 1/√2.So, theta = 45 degrees.So, angle DAB is 45 degrees.Let me verify if all the triangles are indeed isosceles.Triangle DEC: Points D(a,a), E(3a/2, a/2), C(2a,a).Compute DE, EC, DC.DE: distance between D(a,a) and E(3a/2, a/2):sqrt[(3a/2 - a)^2 + (a/2 - a)^2] = sqrt[(a/2)^2 + (-a/2)^2] = sqrt(a²/4 + a²/4) = sqrt(a²/2) = a/√2.EC: distance between E(3a/2, a/2) and C(2a,a):sqrt[(2a - 3a/2)^2 + (a - a/2)^2] = sqrt[(a/2)^2 + (a/2)^2] = sqrt(a²/4 + a²/4) = sqrt(a²/2) = a/√2.DC: distance between D(a,a) and C(2a,a):sqrt[(2a - a)^2 + (a - a)^2] = sqrt[a² + 0] = a.So, DE = EC = a/√2, which is less than DC = a. So, triangle DEC is isosceles with DE = EC.Triangle BED: Points B(a,0), E(3a/2, a/2), D(a,a).Compute BE, ED, BD.BE: distance between B(a,0) and E(3a/2, a/2):sqrt[(3a/2 - a)^2 + (a/2 - 0)^2] = sqrt[(a/2)^2 + (a/2)^2] = sqrt(a²/4 + a²/4) = sqrt(a²/2) = a/√2.ED: distance between E(3a/2, a/2) and D(a,a):sqrt[(a - 3a/2)^2 + (a - a/2)^2] = sqrt[(-a/2)^2 + (a/2)^2] = sqrt(a²/4 + a²/4) = sqrt(a²/2) = a/√2.BD: distance between B(a,0) and D(a,a):sqrt[(a - a)^2 + (a - 0)^2] = sqrt[0 + a²] = a.So, BE = ED = a/√2, which is less than BD = a. So, triangle BED is isosceles with BE = ED.Triangle BAD: Points B(a,0), A(0,0), D(a,a).Compute BA, AD, BD.BA: distance between B(a,0) and A(0,0): a.AD: distance between A(0,0) and D(a,a): sqrt(a² + a²) = a√2.BD: distance between B(a,0) and D(a,a): a.So, BA = BD = a, which makes triangle BAD isosceles with BA = BD.So, all conditions are satisfied when angle DAB is 45 degrees.Wait, but earlier I considered the case where triangle BAD is isosceles with BA = BD, leading to a = c. But what if triangle BAD is isosceles with BA = AD? That would mean a = sqrt(a² + c²), which is only possible if c = 0, which is degenerate. So, that case is invalid.Alternatively, if triangle BAD is isosceles with AD = BD, then sqrt(a² + c²) = c, which implies a = 0, which is also degenerate. So, the only valid case is BA = BD, leading to a = c, and angle DAB = 45 degrees.Therefore, the possible value for angle DAB is 45 degrees.

💡Okay, so I have this problem here: In triangle ABC, if (cos A + sin A)(cos B + sin B) equals 2, then what is angle C? Hmm, interesting. I need to figure out the measure of angle C given this equation. Let me try to break it down step by step.First, I remember that in any triangle, the sum of the angles is 180 degrees. So, A + B + C = 180°. That means if I can find angles A and B, I can easily find angle C by subtracting their sum from 180°. So, my main goal is to find A and B such that (cos A + sin A)(cos B + sin B) = 2.Let me write down the given equation:(cos A + sin A)(cos B + sin B) = 2.I think I can expand this product to make it easier to handle. Let's try that:cos A * cos B + cos A * sin B + sin A * cos B + sin A * sin B = 2.Hmm, that looks a bit complicated, but maybe I can recognize some trigonometric identities here. I remember that cos(A - B) = cos A cos B + sin A sin B. So, the first and last terms in the expansion give me cos(A - B). That's helpful. Let me rewrite the equation:cos(A - B) + cos A sin B + sin A cos B = 2.Wait, the remaining terms, cos A sin B + sin A cos B, that's another identity. I think that's sin(A + B). Let me check: sin(A + B) = sin A cos B + cos A sin B. Yes, that's correct. So, replacing that, the equation becomes:cos(A - B) + sin(A + B) = 2.Alright, so now I have cos(A - B) + sin(A + B) = 2. Hmm, both cosine and sine functions have maximum values of 1. So, for their sum to be 2, each of them must individually be equal to 1. Because if either one is less than 1, the sum would be less than 2. So, that gives me two equations:1. cos(A - B) = 12. sin(A + B) = 1Let me solve each of these equations separately.Starting with the first equation: cos(A - B) = 1. The cosine of an angle is 1 when the angle is 0°, 360°, 720°, etc. But since A and B are angles in a triangle, they must each be between 0° and 180°, and their difference can't be more than 180°, so the only possibility here is that A - B = 0°, meaning A = B.Okay, so angles A and B are equal. That means triangle ABC is isosceles with sides opposite angles A and B being equal. But let's not get ahead of ourselves yet. Let's look at the second equation: sin(A + B) = 1.The sine of an angle is 1 at 90°, 450°, etc. Again, since A and B are angles in a triangle, their sum must be less than 180°, so the only possibility is A + B = 90°.Wait, hold on. If A + B = 90°, and since A = B, then each of them must be 45°, right? Because 45° + 45° = 90°. So, angles A and B are both 45°, which makes angle C equal to 180° - 90° = 90°. So, angle C is 90°, making triangle ABC a right-angled isosceles triangle.But hold on, let me double-check my reasoning because sometimes when dealing with trigonometric identities, especially with multiple angles, things can get tricky.So, starting again, we had:(cos A + sin A)(cos B + sin B) = 2.We expanded it to:cos(A - B) + sin(A + B) = 2.Since both cos(A - B) and sin(A + B) can't exceed 1, their sum can't exceed 2. So, for their sum to be exactly 2, each must individually be 1. So, cos(A - B) = 1 and sin(A + B) = 1.From cos(A - B) = 1, we concluded A = B. From sin(A + B) = 1, we concluded A + B = 90°, so each is 45°, making C = 90°. That seems consistent.Wait, but let me think about the range of angles in a triangle. Each angle must be greater than 0° and less than 180°, and their sum must be exactly 180°. So, if A and B are each 45°, then C is 90°, which is valid because 45° + 45° + 90° = 180°. So, that's good.Is there any other possibility? For example, could A - B be 360°, but that's not possible because angles in a triangle are less than 180°, so their difference can't be 360°. Similarly, sin(A + B) = 1 could be at 450°, but that's way beyond the sum of two angles in a triangle, which is at most 180°. So, the only valid solution is A = B = 45°, and C = 90°.Therefore, angle C is 90°, which is a right angle. So, triangle ABC is a right-angled isosceles triangle.Wait, but the original problem didn't specify whether it's a right triangle or not. It just gave that equation. So, through solving, we found that it must be a right-angled isosceles triangle. So, that makes sense.Let me recap:1. Start with (cos A + sin A)(cos B + sin B) = 2.2. Expand to get cos(A - B) + sin(A + B) = 2.3. Since both cos and sin terms can't exceed 1, each must be 1.4. Therefore, A - B = 0°, so A = B, and A + B = 90°, so A = B = 45°.5. Thus, angle C = 180° - 90° = 90°.Yes, that seems solid. I don't see any mistakes in the reasoning. So, angle C is 90°, which is 90 degrees.But wait, the problem didn't specify whether the triangle is acute, obtuse, or right-angled. So, by solving, we found that it must be right-angled. That's interesting.Alternatively, is there another way to approach this problem? Maybe using other trigonometric identities or perhaps by squaring both sides or something like that? Let me try another method to confirm.Another approach: Let's consider the expression (cos A + sin A). I know that cos A + sin A can be rewritten using the identity:cos A + sin A = √2 sin(A + 45°).Similarly, cos B + sin B = √2 sin(B + 45°).So, substituting these into the original equation:(√2 sin(A + 45°))(√2 sin(B + 45°)) = 2.Multiplying the constants:√2 * √2 = 2, so 2 sin(A + 45°) sin(B + 45°) = 2.Divide both sides by 2:sin(A + 45°) sin(B + 45°) = 1.Hmm, so the product of two sine functions equals 1. Since the maximum value of sine is 1, the only way their product is 1 is if each sine term is 1. So,sin(A + 45°) = 1 and sin(B + 45°) = 1.Which implies:A + 45° = 90° + 360°k and B + 45° = 90° + 360°m, where k and m are integers.But since A and B are angles in a triangle, they must be between 0° and 180°, so the only possible solution is:A + 45° = 90° => A = 45°,and similarly,B + 45° = 90° => B = 45°.Therefore, A = B = 45°, so angle C = 180° - 45° - 45° = 90°.So, this method also leads us to the same conclusion: angle C is 90°. That reinforces my previous result.Another way to think about it: Maybe using vectors or complex numbers? Hmm, that might be overcomplicating it. The two methods I've tried so far both lead to the same answer, so I feel confident that angle C is 90°.Wait, just to be thorough, let me plug the values back into the original equation to verify.If A = 45°, B = 45°, then:cos 45° + sin 45° = √2/2 + √2/2 = √2.Similarly, cos 45° + sin 45° = √2.So, (√2)(√2) = 2, which matches the given equation. Perfect, that checks out.So, everything seems consistent. I don't see any alternative solutions or possibilities here. The only way for the product (cos A + sin A)(cos B + sin B) to equal 2 is if both A and B are 45°, making angle C 90°.Therefore, the measure of angle C is 90 degrees.**Final Answer**The measure of angle ( C ) is boxed{90^circ}.