💡First, I need to determine the total number of books Rahim purchased. He bought 40 books from one shop and 20 books from another shop. Adding these together gives a total of 60 books.Next, I will calculate the total amount of money Rahim spent. He spent Rs. 600 at the first shop and Rs. 240 at the second shop. Adding these amounts results in a total expenditure of Rs. 840.Finally, to find the average price per book, I will divide the total amount spent by the total number of books. Dividing Rs. 840 by 60 books gives an average price of Rs. 14 per book.

💡First, I recognize that Summer has a total of 150 degrees.Since Summer has five more degrees than Jolly, I can determine Jolly's degrees by subtracting 5 from Summer's total.So, Jolly has 150 - 5 = 145 degrees.To find the combined number of degrees they both have, I add Summer's degrees to Jolly's degrees.Therefore, the combined total is 150 + 145 = 295 degrees.

💡Alright, so I have this problem about a sequence {a_n} where the sum S_n is given by S_n = 2a_n - n for all positive integers n. There are two parts: first, I need to prove that the sequence {a_n + 1} is a geometric sequence, and second, I have to find the sum of the odd-indexed terms up to a_{2n+1}. Let me try to tackle this step by step.Starting with part (1). I know that S_n is the sum of the first n terms of the sequence {a_n}. So, S_n = a_1 + a_2 + ... + a_n. But the problem gives me S_n = 2a_n - n. Hmm, that's interesting. So, the sum of the first n terms is related to the nth term itself. I need to find a relationship between a_n and a_{n-1} to see if {a_n + 1} is geometric.First, let's find S_1. For n=1, S_1 = a_1. According to the given formula, S_1 = 2a_1 - 1. So, setting them equal: a_1 = 2a_1 - 1. Solving for a_1, subtract a_1 from both sides: 0 = a_1 - 1, so a_1 = 1. Got that.Now, for n ≥ 2, S_n = 2a_n - n and S_{n-1} = 2a_{n-1} - (n-1). Since S_n - S_{n-1} = a_n, I can write:a_n = S_n - S_{n-1} = (2a_n - n) - (2a_{n-1} - (n - 1)).Let me simplify that:a_n = 2a_n - n - 2a_{n-1} + n - 1.Simplify the terms:a_n = 2a_n - 2a_{n-1} -1.Now, subtract 2a_n from both sides:a_n - 2a_n = -2a_{n-1} -1.Which simplifies to:- a_n = -2a_{n-1} -1.Multiply both sides by -1:a_n = 2a_{n-1} + 1.Okay, so we have a recurrence relation: a_n = 2a_{n-1} + 1 for n ≥ 2, with a_1 = 1.Now, the problem asks to prove that {a_n + 1} is a geometric sequence. Let me define b_n = a_n + 1. Then, let's see what the recurrence becomes.From a_n = 2a_{n-1} + 1, adding 1 to both sides:a_n + 1 = 2a_{n-1} + 2.Which can be written as:b_n = 2(a_{n-1} + 1) = 2b_{n-1}.So, b_n = 2b_{n-1}, which is a geometric sequence with common ratio 2. Also, let's check the first term. Since a_1 = 1, b_1 = a_1 + 1 = 2. So, the first term is 2 and the common ratio is 2. Therefore, {b_n} is a geometric sequence. That proves part (1).Moving on to part (2). I need to find the sum a_1 + a_3 + a_5 + ... + a_{2n+1}. So, it's the sum of the odd-indexed terms up to the (2n+1)th term. Let's see.First, from part (1), we have b_n = a_n + 1 = 2^n, since it's a geometric sequence starting at 2 with ratio 2. Therefore, a_n = 2^n - 1.So, each term a_k = 2^k - 1. Therefore, the sum we need is:Sum = a_1 + a_3 + a_5 + ... + a_{2n+1} = (2^1 - 1) + (2^3 - 1) + (2^5 - 1) + ... + (2^{2n+1} - 1).Let me separate the sum into two parts: the sum of the powers of 2 and the sum of the constants.Sum = (2^1 + 2^3 + 2^5 + ... + 2^{2n+1}) - (1 + 1 + 1 + ... + 1).How many terms are there in each part? Let's see. The exponents are 1, 3, 5, ..., 2n+1. That's an arithmetic sequence with first term 1, last term 2n+1, and common difference 2. The number of terms is ((2n+1) - 1)/2 + 1 = (2n)/2 + 1 = n + 1. So, there are n+1 terms in each part.Therefore, the sum becomes:Sum = (2^1 + 2^3 + 2^5 + ... + 2^{2n+1}) - (n + 1).Now, let's compute the sum of the geometric series: 2^1 + 2^3 + 2^5 + ... + 2^{2n+1}.This is a geometric series with first term 2, common ratio 2^2 = 4, and number of terms n+1.The formula for the sum of a geometric series is S = a*(r^{k} - 1)/(r - 1), where a is the first term, r is the common ratio, and k is the number of terms.So, plugging in the values:Sum_geo = 2*(4^{n+1} - 1)/(4 - 1) = 2*(4^{n+1} - 1)/3.Simplify that:Sum_geo = (2*4^{n+1} - 2)/3.But 4^{n+1} is equal to 2^{2(n+1)} = 2^{2n+2}, so:Sum_geo = (2^{2n+3} - 2)/3.Wait, let me check that. 4^{n+1} is 2^{2(n+1)} = 2^{2n+2}, so 2*4^{n+1} = 2^{2n+3}. So, yes, that part is correct.Therefore, the sum_geo is (2^{2n+3} - 2)/3.Now, subtracting the sum of the constants, which is (n + 1):Sum_total = (2^{2n+3} - 2)/3 - (n + 1).Let me write that as:Sum_total = (2^{2n+3} - 2 - 3(n + 1))/3.Simplify the numerator:2^{2n+3} - 2 - 3n - 3 = 2^{2n+3} - 3n - 5.So, Sum_total = (2^{2n+3} - 3n - 5)/3.Therefore, the sum a_1 + a_3 + a_5 + ... + a_{2n+1} is equal to (2^{2n+3} - 3n - 5)/3.Let me double-check my steps to make sure I didn't make a mistake.1. Found a_1 = 1 correctly.2. Derived the recurrence a_n = 2a_{n-1} + 1 correctly.3. Defined b_n = a_n + 1, leading to b_n = 2b_{n-1}, which is geometric with ratio 2 and first term 2. So, b_n = 2^n, hence a_n = 2^n -1. That seems right.4. For the sum, identified that it's the sum of a_1, a_3, ..., a_{2n+1}, which translates to summing 2^1 -1, 2^3 -1, ..., 2^{2n+1} -1. That makes sense.5. Calculated the number of terms as n+1, which is correct because starting from 1 to 2n+1 with step 2 gives (2n+1 -1)/2 +1 = n+1.6. Summed the geometric series correctly: first term 2, ratio 4, n+1 terms. Applied the formula correctly, leading to (2^{2n+3} - 2)/3.7. Subtracted (n +1), converted to a common denominator, leading to (2^{2n+3} -3n -5)/3. That seems correct.I think all steps are valid, so the final expression should be correct.

💡Alright, so I have this geometry problem here about a quadrilateral ABCD. The problem says that the perpendicular bisectors of sides AB and CD intersect on side AD. Also, it's given that angle A is equal to angle D. I need to prove that the diagonals of quadrilateral ABCD are equal. Hmm, okay, let me try to visualize this.First, let me recall what a perpendicular bisector is. It's a line that is perpendicular to a side of the quadrilateral and passes through its midpoint. So, if I have quadrilateral ABCD, the perpendicular bisector of AB would pass through the midpoint of AB and be perpendicular to it. Similarly, the perpendicular bisector of CD would pass through the midpoint of CD and be perpendicular to it.Now, these two perpendicular bisectors intersect on side AD. Let me denote the point of intersection as E. So, E is on AD, and it's the intersection point of the perpendicular bisectors of AB and CD. That means E is equidistant from A and B because it's on the perpendicular bisector of AB. Similarly, E is equidistant from C and D because it's on the perpendicular bisector of CD.So, from this, I can write that EA = EB and EC = ED. That's because any point on the perpendicular bisector of a segment is equidistant from the endpoints of that segment.Given that angle A is equal to angle D, which is ∠A = ∠D. I need to use this information somehow to show that the diagonals AC and BD are equal.Let me try to draw this quadrilateral. I have ABCD with AB, BC, CD, DA as sides. The perpendicular bisectors of AB and CD intersect at point E on AD. So, E is somewhere along AD, and it's equidistant from A and B, and also from C and D.Since EA = EB and EC = ED, triangles EAB and ECD are isosceles. In triangle EAB, sides EA and EB are equal, so the base angles at A and B are equal. Similarly, in triangle ECD, sides EC and ED are equal, so the base angles at C and D are equal.Given that ∠A = ∠D, and since in triangle EAB, ∠EAB = ∠EBA, and in triangle ECD, ∠EDC = ∠ECD, maybe there's a relationship between these angles that can help me.Let me denote some angles. Let’s say ∠EAB = ∠EBA = x, and ∠EDC = ∠ECD = y. Since ∠A = ∠D, and ∠A is at vertex A, which is angle ∠DAB, and ∠D is at vertex D, which is angle ∠ADC. So, ∠DAB = ∠ADC.But in triangle EAB, ∠EAB = x, and in triangle ECD, ∠EDC = y. Hmm, not sure if that's directly helpful yet.Wait, maybe I can relate the angles at point E. Since E is the intersection of the perpendicular bisectors, maybe there are some vertical angles or something that can be used.Let me think about the triangles involved. Since EA = EB and EC = ED, maybe triangles EAC and EBD are congruent or something like that. If I can show that triangles EAC and EBD are congruent, then AC = BD, which would be the diagonals.To show that triangles EAC and EBD are congruent, I need to have some corresponding sides and angles equal. I know EA = EB and EC = ED. So, sides EA and EB are equal, and sides EC and ED are equal. Also, if I can find an angle between these sides that's equal, maybe I can use SAS or ASA congruence.Looking at angles at point E, ∠AEC and ∠BED. If I can show that these angles are equal, then with EA = EB and EC = ED, triangles EAC and EBD would be congruent by SAS.So, how can I show that ∠AEC = ∠BED? Let me think about the angles around point E.Since E is on AD, and it's the intersection of the perpendicular bisectors, the lines from E to A and E to D are just parts of AD. The other lines from E to B and E to C are the perpendicular bisectors.Wait, maybe I can use the fact that the perpendicular bisectors are perpendicular to AB and CD. So, the angles at E with respect to AB and CD are right angles.Let me denote the midpoints of AB and CD as M and N, respectively. So, EM is the perpendicular bisector of AB, and EN is the perpendicular bisector of CD. Therefore, EM ⊥ AB and EN ⊥ CD.Since EM and EN are perpendicular to AB and CD, respectively, and E is their intersection on AD, maybe there are some right triangles involved here.Looking at triangles EMA and ENB, but I'm not sure if that's helpful. Maybe I should consider the angles at E.Alternatively, since EA = EB and EC = ED, maybe I can consider the triangles EAB and ECD. They are both isosceles, and if I can relate their base angles, maybe I can find some relationship.Given that ∠A = ∠D, and in triangle EAB, ∠EAB = x, and in triangle ECD, ∠EDC = y, and ∠A = ∠D, which are angles at A and D. So, ∠DAB = ∠ADC.But ∠DAB is part of triangle EAB, and ∠ADC is part of triangle ECD. Maybe I can express these angles in terms of x and y.Wait, in triangle EAB, ∠EAB = x, and the other base angle is also x, so the vertex angle at E is 180 - 2x. Similarly, in triangle ECD, the vertex angle at E is 180 - 2y.But since ∠A = ∠D, which are angles at A and D, maybe I can relate x and y somehow.Wait, ∠A is ∠DAB, which is the angle between sides AB and AD. Similarly, ∠D is ∠ADC, which is the angle between sides DC and DA.Since E is on AD, maybe I can express ∠DAB and ∠ADC in terms of angles involving E.Let me consider triangle ABE. Since EA = EB, it's isosceles, so ∠EAB = ∠EBA = x. Similarly, in triangle CDE, EC = ED, so ∠ECD = ∠EDC = y.Now, ∠DAB = ∠EAB + ∠EAD. But since E is on AD, ∠EAD is just a straight line, so ∠EAD = 0? Wait, no, that doesn't make sense.Wait, maybe I'm complicating things. Let me try a different approach.Since E is on AD, and EA = EB, EC = ED, maybe I can consider the power of point E with respect to the circumcircles of triangles ABC and ADC or something like that. But I'm not sure if that's necessary.Alternatively, maybe I can use coordinate geometry. Let me assign coordinates to the quadrilateral and see if I can derive the result.Let me place point A at (0,0) and point D at (d,0) on the x-axis. Since E is on AD, let me denote E as (e,0), where 0 < e < d.Since E is the intersection of the perpendicular bisectors of AB and CD, let me denote the midpoints of AB and CD as M and N, respectively.Let me denote point B as (b_x, b_y) and point C as (c_x, c_y). Then, the midpoint M of AB is ((b_x)/2, (b_y)/2), and the midpoint N of CD is ((d + c_x)/2, (0 + c_y)/2).The perpendicular bisector of AB passes through M and is perpendicular to AB. Similarly, the perpendicular bisector of CD passes through N and is perpendicular to CD.Since E is the intersection of these two perpendicular bisectors, it must satisfy the equations of both perpendicular bisectors.Let me write the equation of the perpendicular bisector of AB. The slope of AB is (b_y - 0)/(b_x - 0) = b_y / b_x. Therefore, the slope of the perpendicular bisector is -b_x / b_y.So, the equation of the perpendicular bisector of AB is:(y - (b_y)/2) = (-b_x / b_y)(x - (b_x)/2)Similarly, the slope of CD is (c_y - 0)/(c_x - d) = c_y / (c_x - d). Therefore, the slope of the perpendicular bisector of CD is -(c_x - d)/c_y.The equation of the perpendicular bisector of CD is:(y - (c_y)/2) = [-(c_x - d)/c_y](x - (d + c_x)/2)Since E is (e,0), it must satisfy both equations.Plugging E into the first equation:0 - (b_y)/2 = (-b_x / b_y)(e - (b_x)/2)=> - (b_y)/2 = (-b_x / b_y)(e - (b_x)/2)Multiply both sides by b_y:- (b_y)^2 / 2 = -b_x (e - (b_x)/2)Multiply both sides by -1:(b_y)^2 / 2 = b_x (e - (b_x)/2)=> (b_y)^2 / 2 = b_x e - (b_x)^2 / 2Multiply both sides by 2:(b_y)^2 = 2 b_x e - (b_x)^2=> (b_x)^2 + (b_y)^2 = 2 b_x eSimilarly, plugging E into the second equation:0 - (c_y)/2 = [-(c_x - d)/c_y](e - (d + c_x)/2)=> - (c_y)/2 = [-(c_x - d)/c_y](e - (d + c_x)/2)Multiply both sides by c_y:- (c_y)^2 / 2 = -(c_x - d)(e - (d + c_x)/2)Multiply both sides by -1:(c_y)^2 / 2 = (c_x - d)(e - (d + c_x)/2)Let me simplify the right side:(c_x - d)(e - (d + c_x)/2) = (c_x - d)( (2e - d - c_x)/2 ) = (c_x - d)(2e - d - c_x)/2So, we have:(c_y)^2 / 2 = (c_x - d)(2e - d - c_x)/2Multiply both sides by 2:(c_y)^2 = (c_x - d)(2e - d - c_x)Now, we have two equations:1. (b_x)^2 + (b_y)^2 = 2 b_x e2. (c_y)^2 = (c_x - d)(2e - d - c_x)These equations relate the coordinates of points B and C with the parameter e, which is the x-coordinate of point E on AD.Now, given that angle A is equal to angle D, which is ∠DAB = ∠ADC.Let me express these angles in terms of coordinates.∠DAB is the angle at point A between sides AB and AD. Since AD is along the x-axis from (0,0) to (d,0), the direction of AD is along the positive x-axis. The direction of AB is from A(0,0) to B(b_x, b_y). So, the angle ∠DAB is the angle between vector AB and the x-axis.Similarly, ∠ADC is the angle at point D between sides DC and DA. DA is from D(d,0) to A(0,0), which is along the negative x-axis. DC is from D(d,0) to C(c_x, c_y). So, the angle ∠ADC is the angle between vector DC and the negative x-axis.Given that ∠DAB = ∠ADC, let's express these angles using slopes or vectors.The slope of AB is b_y / b_x, so the angle ∠DAB is arctan(b_y / b_x).The slope of DC is (c_y - 0)/(c_x - d) = c_y / (c_x - d). Since ∠ADC is the angle between DC and the negative x-axis, it's equal to arctan(c_y / (c_x - d)).Given that ∠DAB = ∠ADC, we have:arctan(b_y / b_x) = arctan(c_y / (c_x - d))Which implies:b_y / b_x = c_y / (c_x - d)=> b_y (c_x - d) = b_x c_ySo, that's another equation relating the coordinates.Now, let me summarize the equations I have:1. (b_x)^2 + (b_y)^2 = 2 b_x e2. (c_y)^2 = (c_x - d)(2e - d - c_x)3. b_y (c_x - d) = b_x c_yI need to find a relationship between the diagonals AC and BD. The diagonals are AC from A(0,0) to C(c_x, c_y) and BD from B(b_x, b_y) to D(d,0). The lengths of these diagonals are:AC = sqrt( (c_x)^2 + (c_y)^2 )BD = sqrt( (d - b_x)^2 + (0 - b_y)^2 ) = sqrt( (d - b_x)^2 + (b_y)^2 )I need to show that AC = BD, which would mean:sqrt( (c_x)^2 + (c_y)^2 ) = sqrt( (d - b_x)^2 + (b_y)^2 )Squaring both sides:(c_x)^2 + (c_y)^2 = (d - b_x)^2 + (b_y)^2So, I need to show that:(c_x)^2 + (c_y)^2 = d^2 - 2 d b_x + (b_x)^2 + (b_y)^2From equation 1, I have (b_x)^2 + (b_y)^2 = 2 b_x e. Let me substitute that into the equation:(c_x)^2 + (c_y)^2 = d^2 - 2 d b_x + 2 b_x eSo, I need to show that:(c_x)^2 + (c_y)^2 = d^2 - 2 d b_x + 2 b_x eLet me see if I can express (c_x)^2 + (c_y)^2 in terms of other variables.From equation 2, (c_y)^2 = (c_x - d)(2e - d - c_x)Let me expand the right side:(c_x - d)(2e - d - c_x) = (c_x - d)(-c_x + 2e - d) = -c_x^2 + 2e c_x - d c_x + d c_x - 2e d + d^2Simplify:- c_x^2 + 2e c_x - d c_x + d c_x - 2e d + d^2 = -c_x^2 + 2e c_x - 2e d + d^2So, (c_y)^2 = -c_x^2 + 2e c_x - 2e d + d^2Therefore, (c_x)^2 + (c_y)^2 = (c_x)^2 + (-c_x^2 + 2e c_x - 2e d + d^2) = 2e c_x - 2e d + d^2So, (c_x)^2 + (c_y)^2 = 2e c_x - 2e d + d^2Now, let's go back to the equation we needed to show:(c_x)^2 + (c_y)^2 = d^2 - 2 d b_x + 2 b_x eFrom above, (c_x)^2 + (c_y)^2 = 2e c_x - 2e d + d^2So, we have:2e c_x - 2e d + d^2 = d^2 - 2 d b_x + 2 b_x eSimplify both sides:Left side: 2e c_x - 2e d + d^2Right side: d^2 - 2 d b_x + 2 b_x eSubtract d^2 from both sides:2e c_x - 2e d = -2 d b_x + 2 b_x eLet me factor out 2e on the left and 2b_x on the right:2e (c_x - d) = 2 b_x (e - d)Divide both sides by 2:e (c_x - d) = b_x (e - d)From equation 3, we have b_y (c_x - d) = b_x c_yLet me solve equation 3 for (c_x - d):(c_x - d) = (b_x c_y) / b_ySubstitute this into the equation e (c_x - d) = b_x (e - d):e * (b_x c_y / b_y) = b_x (e - d)Divide both sides by b_x (assuming b_x ≠ 0):e * (c_y / b_y) = e - dFrom equation 1: (b_x)^2 + (b_y)^2 = 2 b_x eLet me solve for e:e = ( (b_x)^2 + (b_y)^2 ) / (2 b_x )So, e = (b_x)/2 + (b_y)^2 / (2 b_x )Let me substitute e into the equation e * (c_y / b_y) = e - d:[ (b_x)/2 + (b_y)^2 / (2 b_x ) ] * (c_y / b_y ) = [ (b_x)/2 + (b_y)^2 / (2 b_x ) ] - dLet me denote this as:Left side: [ (b_x)/2 + (b_y)^2 / (2 b_x ) ] * (c_y / b_y )Right side: [ (b_x)/2 + (b_y)^2 / (2 b_x ) ] - dThis seems complicated, but maybe I can find a relationship between c_y and b_y.From equation 3: b_y (c_x - d) = b_x c_yFrom equation 2: (c_y)^2 = (c_x - d)(2e - d - c_x)Let me substitute (c_x - d) from equation 3 into equation 2:(c_y)^2 = (b_x c_y / b_y ) (2e - d - c_x )So,(c_y)^2 = (b_x c_y / b_y ) (2e - d - c_x )Divide both sides by c_y (assuming c_y ≠ 0):c_y = (b_x / b_y ) (2e - d - c_x )So,c_y = (b_x / b_y ) (2e - d - c_x )Let me solve for c_x:c_y = (b_x / b_y ) (2e - d - c_x )Multiply both sides by b_y:b_y c_y = b_x (2e - d - c_x )From equation 3: b_y (c_x - d) = b_x c_ySo, b_y c_y = b_x (c_x - d )Wait, but from above, b_y c_y = b_x (2e - d - c_x )Therefore,b_x (c_x - d ) = b_x (2e - d - c_x )Assuming b_x ≠ 0, we can divide both sides by b_x:c_x - d = 2e - d - c_xSimplify:c_x - d = 2e - d - c_xAdd d to both sides:c_x = 2e - c_xAdd c_x to both sides:2 c_x = 2eDivide by 2:c_x = eSo, c_x = eThat's interesting. So, the x-coordinate of point C is equal to e, which is the x-coordinate of point E on AD.Now, since c_x = e, let's substitute this back into equation 3:b_y (c_x - d) = b_x c_y=> b_y (e - d) = b_x c_ySo,c_y = (b_y (e - d )) / b_xNow, let's substitute c_x = e into equation 2:(c_y)^2 = (c_x - d)(2e - d - c_x )=> (c_y)^2 = (e - d)(2e - d - e )Simplify:(2e - d - e ) = (e - d )So,(c_y)^2 = (e - d)(e - d ) = (e - d )^2Therefore,c_y = ±(e - d )But since c_y is a coordinate, it can be positive or negative, but let's consider the positive case first.So, c_y = e - d or c_y = d - e, depending on the orientation.But from equation 3, c_y = (b_y (e - d )) / b_xSo, if c_y = e - d, then:e - d = (b_y (e - d )) / b_xAssuming e ≠ d (since E is on AD, and if e = d, E would coincide with D, but then the perpendicular bisector of CD would be undefined as CD would have zero length, which isn't the case), we can divide both sides by (e - d ):1 = b_y / b_xSo,b_y = b_xSimilarly, if c_y = d - e, then:d - e = (b_y (e - d )) / b_xMultiply both sides by b_x:b_x (d - e ) = b_y (e - d )=> b_x (d - e ) = -b_y (d - e )Assuming d ≠ e, we can divide both sides by (d - e ):b_x = -b_ySo, either b_y = b_x or b_y = -b_xBut let's think about the geometry. If b_y = b_x, then point B is at (b_x, b_x), which would mean AB is at a 45-degree angle from the x-axis. Similarly, if b_y = -b_x, then point B is at (b_x, -b_x), which would mean AB is at a -45-degree angle from the x-axis.But since E is on AD, which is along the x-axis, and the perpendicular bisectors intersect at E, the position of B affects the angles.Given that angle A = angle D, which are both angles between sides and AD, it's likely that the slopes of AB and DC are related in a way that their angles with AD are equal.From earlier, we have that the slopes of AB and DC are related by b_y / b_x = c_y / (c_x - d )But since c_x = e and c_y = e - d or d - e, let's substitute:Case 1: c_y = e - dThen,b_y / b_x = (e - d ) / (e - d ) = 1So, b_y / b_x = 1 => b_y = b_xCase 2: c_y = d - eThen,b_y / b_x = (d - e ) / (e - d ) = -1So, b_y / b_x = -1 => b_y = -b_xSo, in either case, b_y = ±b_xNow, let's consider the diagonals AC and BD.AC is from A(0,0) to C(e, c_y). So, AC = sqrt(e^2 + c_y^2 )BD is from B(b_x, b_y) to D(d,0). So, BD = sqrt( (d - b_x)^2 + (0 - b_y)^2 ) = sqrt( (d - b_x)^2 + b_y^2 )We need to show that AC = BD.So,sqrt(e^2 + c_y^2 ) = sqrt( (d - b_x)^2 + b_y^2 )Squaring both sides:e^2 + c_y^2 = (d - b_x)^2 + b_y^2From earlier, c_y = e - d or c_y = d - e, so c_y^2 = (e - d )^2So,e^2 + (e - d )^2 = (d - b_x)^2 + b_y^2Expand both sides:Left side: e^2 + (e^2 - 2 e d + d^2 ) = 2 e^2 - 2 e d + d^2Right side: (d - b_x)^2 + b_y^2 = d^2 - 2 d b_x + b_x^2 + b_y^2From equation 1: (b_x)^2 + (b_y)^2 = 2 b_x eSo, substitute into the right side:d^2 - 2 d b_x + 2 b_x eTherefore, we have:Left side: 2 e^2 - 2 e d + d^2Right side: d^2 - 2 d b_x + 2 b_x eSo,2 e^2 - 2 e d + d^2 = d^2 - 2 d b_x + 2 b_x eSubtract d^2 from both sides:2 e^2 - 2 e d = -2 d b_x + 2 b_x eFactor out 2e on the left and 2b_x on the right:2e (e - d ) = 2 b_x (e - d )Assuming e ≠ d (since E is on AD but not coinciding with D), we can divide both sides by 2(e - d ):e = b_xSo, e = b_xBut from equation 1: (b_x)^2 + (b_y)^2 = 2 b_x eSince e = b_x, substitute:(b_x)^2 + (b_y)^2 = 2 b_x * b_x = 2 (b_x)^2So,(b_x)^2 + (b_y)^2 = 2 (b_x)^2Subtract (b_x)^2 from both sides:(b_y)^2 = (b_x)^2So,b_y = ±b_xWhich matches our earlier conclusion.Therefore, in either case, whether b_y = b_x or b_y = -b_x, we have that AC = BD.Hence, the diagonals AC and BD are equal.**Final Answer**boxed{AC = BD}

💡Alright, so I'm trying to solve this geometry problem about a parallelogram. Let me read it again to make sure I understand all the parts.We have a parallelogram ABCD. Points E and F are on sides AD and CD respectively, such that angle BCE is equal to angle BAF. Then, points K and L are on AD and CD such that AK equals ED and CL equals FD. We need to prove that angle BKD equals angle BLD.Okay, so first, let me visualize the parallelogram. In a parallelogram, opposite sides are equal and parallel. So, AB is parallel to CD, and AD is parallel to BC. Also, AB equals CD, and AD equals BC.Points E and F are on AD and CD. So, E is somewhere between A and D on side AD, and F is somewhere between C and D on side CD. The condition given is that angle BCE equals angle BAF. That seems important. So, angle at point C between BC and CE is equal to the angle at point A between BA and AF.Then, points K and L are defined such that AK equals ED and CL equals FD. So, K is on AD, and since AK equals ED, K is located such that the segment from A to K is equal in length to the segment from E to D. Similarly, L is on CD such that CL equals FD, meaning the segment from C to L is equal in length to the segment from F to D.We need to prove that angle BKD equals angle BLD. So, looking at points B, K, D and B, L, D, we need to show that the angles at K and L are equal.Hmm, okay. Maybe I can draw this out step by step. Since I can't actually draw here, I'll try to imagine it and describe my thoughts.First, let's note that in a parallelogram, opposite angles are equal, and consecutive angles are supplementary. So, angle A equals angle C, and angle B equals angle D. Also, the diagonals bisect each other, but I don't know if that's directly useful here.Given that angle BCE equals angle BAF, maybe there's some similarity or congruence we can exploit. Let me denote angle BCE and angle BAF as some angle alpha. So, angle BCE = angle BAF = alpha.Since ABCD is a parallelogram, AB is parallel to CD, and AD is parallel to BC. So, maybe there are some alternate interior angles or corresponding angles we can use.Let me consider triangle BCE and triangle BAF. If angle BCE equals angle BAF, and if we can find another pair of equal angles or sides, maybe these triangles are similar.Wait, triangle BCE has sides BC, CE, and BE, and triangle BAF has sides BA, AF, and BF. Hmm, not sure if they are similar yet.Alternatively, maybe there's some reflection or rotation symmetry here. Since ABCD is a parallelogram, it has rotational symmetry of order 2, meaning if you rotate it 180 degrees around the center, it maps onto itself.But I'm not sure if that helps directly. Maybe I should look at the points K and L.Given that AK = ED and CL = FD, perhaps K and L are constructed in a way that relates to E and F. Since AK = ED, and K is on AD, that means that K divides AD into segments AK and KD, where AK is equal to ED. Similarly, L divides CD into CL and LD, where CL is equal to FD.So, if I let the length of ED be x, then AK is also x. Similarly, if FD is y, then CL is also y.Since AD is a side of the parallelogram, its length is equal to BC. Let's denote AD = BC = m, and AB = CD = n.So, on AD, we have point E somewhere, so AE + ED = AD = m. Similarly, on CD, CF + FD = CD = n.Given that AK = ED = x, then KD = AD - AK = m - x.Similarly, CL = FD = y, so LD = CD - CL = n - y.Hmm, okay. Maybe I can express the positions of K and L in terms of E and F.Alternatively, maybe I can use vectors or coordinate geometry. Sometimes, assigning coordinates can make things clearer.Let me try that. Let's place the parallelogram ABCD on a coordinate system. Let me assign coordinates to the vertices.Let’s say point A is at (0, 0). Since it's a parallelogram, let's let point B be at (a, 0), point D at (0, b), so point C would be at (a, b).So, coordinates:- A: (0, 0)- B: (a, 0)- C: (a, b)- D: (0, b)Now, points E and F are on AD and CD respectively.Point E is on AD, which goes from (0, 0) to (0, b). Let's parameterize E. Let’s say E is at (0, e), where 0 < e < b.Similarly, point F is on CD, which goes from (a, b) to (0, b). Wait, no. CD goes from C(a, b) to D(0, b). So, CD is a horizontal line at y = b from x = a to x = 0.Wait, actually, in a parallelogram, CD is from C(a, b) to D(0, b). So, CD is a horizontal line segment.Wait, but in my coordinate system, CD is from (a, b) to (0, b). So, point F is somewhere on CD, so its coordinates can be expressed as (f, b), where 0 < f < a.Similarly, point E is on AD, which is from (0, 0) to (0, b), so E is at (0, e), as I said.Now, the condition is that angle BCE equals angle BAF.Let me compute these angles in terms of coordinates.First, angle BCE is the angle at point C between points B, C, and E.Point B is at (a, 0), point C is at (a, b), and point E is at (0, e).So, the vectors CB and CE can be used to find angle BCE.Vector CB is from C to B: (a - a, 0 - b) = (0, -b).Vector CE is from C to E: (0 - a, e - b) = (-a, e - b).Similarly, angle BAF is the angle at point A between points B, A, and F.Point B is at (a, 0), point A is at (0, 0), and point F is at (f, b).So, vectors AB and AF can be used to find angle BAF.Vector AB is from A to B: (a, 0).Vector AF is from A to F: (f, b).So, angle BCE is the angle between vectors CB and CE, and angle BAF is the angle between vectors AB and AF.Given that angle BCE = angle BAF, so the angle between CB and CE equals the angle between AB and AF.We can use the dot product to express the cosine of these angles.The cosine of angle between two vectors u and v is (u · v)/(|u||v|).So, for angle BCE:Vectors CB = (0, -b) and CE = (-a, e - b).Dot product CB · CE = (0)(-a) + (-b)(e - b) = -b(e - b) = -be + b².The magnitude of CB is sqrt(0² + (-b)²) = b.The magnitude of CE is sqrt((-a)² + (e - b)²) = sqrt(a² + (e - b)²).So, cos(angle BCE) = (-be + b²)/(b * sqrt(a² + (e - b)²)) = (b² - be)/(b * sqrt(a² + (e - b)²)) = (b - e)/sqrt(a² + (e - b)²).Similarly, for angle BAF:Vectors AB = (a, 0) and AF = (f, b).Dot product AB · AF = a*f + 0*b = af.The magnitude of AB is sqrt(a² + 0²) = a.The magnitude of AF is sqrt(f² + b²).So, cos(angle BAF) = af/(a * sqrt(f² + b²)) = f/sqrt(f² + b²).Given that angle BCE = angle BAF, so their cosines are equal:(b - e)/sqrt(a² + (e - b)²) = f/sqrt(f² + b²).Hmm, that's an equation relating e and f. Maybe I can square both sides to eliminate the square roots.[(b - e)²]/[a² + (e - b)²] = [f²]/[f² + b²].Note that (e - b)² is the same as (b - e)², so let's denote (b - e)² as t.Then, the equation becomes:t / (a² + t) = f² / (f² + b²).Cross-multiplying:t(f² + b²) = f²(a² + t).Expanding both sides:t f² + t b² = f² a² + f² t.Subtract t f² from both sides:t b² = f² a².So, t = (f² a²)/b².But t is (b - e)², so:(b - e)² = (f² a²)/b².Taking square roots:b - e = (f a)/b.So, e = b - (f a)/b.So, e = (b² - a f)/b.Okay, so that relates e and f. So, if I know f, I can find e, or vice versa.Now, moving on to points K and L.Point K is on AD such that AK = ED.Since AD is from (0,0) to (0, b), point K is somewhere on the y-axis between A and D.Given that AK = ED.Point E is at (0, e), so ED is the length from E to D, which is b - e.Therefore, AK = ED = b - e.Since AK is the length from A(0,0) to K(0, k), so AK = k.Therefore, k = b - e.So, point K is at (0, k) = (0, b - e).Similarly, point L is on CD such that CL = FD.Point F is at (f, b), so FD is the length from F to D.Since CD is from (a, b) to (0, b), FD is the length from F(f, b) to D(0, b), which is |f - 0| = f.Wait, no. Wait, CD is from C(a, b) to D(0, b). So, FD is the segment from F to D, which is along CD.Since CD is a horizontal line from (a, b) to (0, b), the length of FD is the distance from F(f, b) to D(0, b), which is |f - 0| = f.Therefore, CL = FD = f.Point L is on CD, which is from C(a, b) to D(0, b). So, CL is the length from C(a, b) to L(l, b), which is |a - l|.Given that CL = f, so |a - l| = f.Since L is between C and D, l is between 0 and a, so a - l = f, so l = a - f.Therefore, point L is at (l, b) = (a - f, b).So, now we have coordinates for K and L.Point K is at (0, b - e).Point L is at (a - f, b).Now, we need to prove that angle BKD equals angle BLD.So, angle at K between points B, K, D and angle at L between points B, L, D.So, let's find the coordinates of points B, K, D, and L.Point B: (a, 0)Point K: (0, b - e)Point D: (0, b)Point L: (a - f, b)So, angle BKD is the angle at K between points B, K, D.Similarly, angle BLD is the angle at L between points B, L, D.To find these angles, we can compute the slopes of the lines or use vectors.Alternatively, we can compute the vectors from K to B and from K to D, and similarly from L to B and from L to D, and then find the angles between these vectors.Let me try that.First, for angle BKD at point K.Vectors from K to B and from K to D.Point K: (0, b - e)Point B: (a, 0)Point D: (0, b)So, vector KB = B - K = (a - 0, 0 - (b - e)) = (a, e - b)Vector KD = D - K = (0 - 0, b - (b - e)) = (0, e)Similarly, for angle BLD at point L.Point L: (a - f, b)Point B: (a, 0)Point D: (0, b)So, vector LB = B - L = (a - (a - f), 0 - b) = (f, -b)Vector LD = D - L = (0 - (a - f), b - b) = (f - a, 0)Now, to find the angles between these vectors, we can use the dot product formula again.For angle BKD, between vectors KB and KD.Vectors KB = (a, e - b) and KD = (0, e).Dot product KB · KD = a*0 + (e - b)*e = 0 + e(e - b) = e² - b e.The magnitude of KB is sqrt(a² + (e - b)²).The magnitude of KD is sqrt(0² + e²) = e.So, cos(angle BKD) = (e² - b e)/(sqrt(a² + (e - b)²) * e) = (e - b)/sqrt(a² + (e - b)²).Similarly, for angle BLD, between vectors LB and LD.Vectors LB = (f, -b) and LD = (f - a, 0).Dot product LB · LD = f*(f - a) + (-b)*0 = f² - a f.The magnitude of LB is sqrt(f² + b²).The magnitude of LD is sqrt((f - a)² + 0²) = |f - a| = a - f (since f < a).So, cos(angle BLD) = (f² - a f)/(sqrt(f² + b²) * (a - f)).Simplify numerator: f² - a f = f(f - a) = -f(a - f).Denominator: sqrt(f² + b²) * (a - f).So, cos(angle BLD) = (-f(a - f))/(sqrt(f² + b²)*(a - f)) = -f / sqrt(f² + b²).But cosine is even, so cos(angle) = |value|, but angles in geometry are between 0 and 180, so the sign might indicate the direction, but the magnitude is what we need.Wait, but in our earlier calculation for angle BKD, we had:cos(angle BKD) = (e - b)/sqrt(a² + (e - b)²).But from earlier, we had that (b - e) = (a f)/b, so e = b - (a f)/b.So, e - b = - (a f)/b.Therefore, cos(angle BKD) = (- (a f)/b)/sqrt(a² + ( (a f)/b )²).Simplify denominator:sqrt(a² + (a² f²)/b²) = a sqrt(1 + (f²)/b²) = a sqrt( (b² + f²)/b² ) = (a / b) sqrt(b² + f²).So, cos(angle BKD) = (- (a f)/b ) / ( (a / b) sqrt(b² + f²) ) = (- f)/sqrt(b² + f²).But cosine is negative, which would imply that the angle is greater than 90 degrees. However, in the context of the problem, angles BKD and BLD are likely to be acute or obtuse depending on the configuration, but we need to compare their cosines.Wait, but in our earlier calculation, cos(angle BKD) = (-f)/sqrt(f² + b²), and cos(angle BLD) = (-f)/sqrt(f² + b²).Wait, that's interesting. Both cosines are equal to (-f)/sqrt(f² + b²). So, cos(angle BKD) = cos(angle BLD).Since both angles are between 0 and 180 degrees, their cosines being equal implies that the angles themselves are equal.Therefore, angle BKD equals angle BLD.So, that's the proof.But let me double-check my steps to make sure I didn't make a mistake.1. Assigned coordinates correctly.2. Expressed points E and F in terms of e and f.3. Used the condition angle BCE = angle BAF to relate e and f, leading to e = b - (a f)/b.4. Found coordinates of K and L based on AK = ED and CL = FD.5. Expressed vectors KB, KD, LB, LD.6. Computed dot products and magnitudes to find cosines of angles BKD and BLD.7. Found that both cosines equal (-f)/sqrt(f² + b²), hence angles are equal.Yes, that seems correct. So, the key was to assign coordinates, express everything in terms of variables, use the given angle condition to relate e and f, then compute the necessary vectors and dot products to find the cosines of the target angles, which turned out to be equal.

💡Okay, so I'm trying to solve this problem where α is given as 1690 degrees. The problem has two parts. First, I need to express α in the form of 2kπ + β, where k is an integer and β is between 0 and 2π. Second, I need to find an angle θ that has the same terminal side as α but lies between -4π and -2π. Alright, starting with the first part. I remember that angles in radians can be converted from degrees by multiplying by π/180. So, 1690 degrees in radians would be 1690 * (π/180). Let me calculate that. 1690 divided by 180 is... let me see, 180 times 9 is 1620, so 1690 minus 1620 is 70. So that's 9 and 70/180, which simplifies to 9 and 7/18. So, 1690 degrees is 9π + (7π/18). Wait, but the question wants it in the form 2kπ + β, where β is between 0 and 2π. So, I need to express 9π + (7π/18) as a multiple of 2π plus some remainder. Let me figure out how many times 2π goes into 9π. 2π times 4 is 8π, and 2π times 5 is 10π. Since 9π is between 8π and 10π, the integer k would be 4. So, 2kπ is 8π, and then the remainder would be 9π - 8π, which is π. But wait, I also have that 7π/18. So, adding that to π, which is 18π/18, gives me (18π + 7π)/18 = 25π/18. So, putting it all together, α is 8π + 25π/18, which is 4*2π + 25π/18. That should be the answer for part (1). Now, moving on to part (2). I need to find θ such that θ has the same terminal side as α and lies between -4π and -2π. Since angles with the same terminal side differ by integer multiples of 2π, θ can be written as α + 2kπ for some integer k. But α is 1690 degrees, which we've already converted to radians as 25π/18 + 8π. So, θ = 25π/18 + 8π + 2kπ. Wait, no, actually, θ should be equal to α + 2kπ, but since α is already expressed as 4*2π + 25π/18, θ can be written as 25π/18 + 2kπ. But we need θ to be between -4π and -2π. So, let's set up the inequality: -4π < 25π/18 + 2kπ < -2π. To solve for k, I can subtract 25π/18 from all parts of the inequality: -4π - 25π/18 < 2kπ < -2π - 25π/18. Let me compute the left side: -4π is -72π/18, so -72π/18 -25π/18 = -97π/18. Similarly, the right side: -2π is -36π/18, so -36π/18 -25π/18 = -61π/18. So, the inequality becomes: -97π/18 < 2kπ < -61π/18. Dividing all parts by 2π: (-97π/18)/(2π) < k < (-61π/18)/(2π). Simplifying, the π cancels out: -97/36 < k < -61/36. Calculating these fractions: -97 divided by 36 is approximately -2.694, and -61 divided by 36 is approximately -1.694. Since k must be an integer, the only integer between -2.694 and -1.694 is k = -2. So, plugging k = -2 back into θ = 25π/18 + 2kπ: θ = 25π/18 + 2*(-2)π = 25π/18 - 4π. Converting 4π to eighteenths: 4π = 72π/18. So, θ = 25π/18 - 72π/18 = -47π/18. Let me check if -47π/18 is indeed between -4π and -2π. -4π is -72π/18, and -2π is -36π/18. -47π/18 is approximately -2.61π, which is between -4π and -2π. So, that works. Therefore, the answers are: (1) α = 4*2π + 25π/18 (2) θ = -47π/18 I think that's it. Let me just double-check my calculations to make sure I didn't make any mistakes. For part (1), converting 1690 degrees to radians: 1690 * π/180 = 169π/18. Then, 169 divided by 18 is 9 with a remainder of 7, so 9π + 7π/18. Since 9π is 4*2π + π, adding 7π/18 gives 4*2π + 25π/18. That seems correct. For part (2), starting from θ = 25π/18 + 2kπ, setting up the inequality, solving for k, and finding k = -2. Then, θ = 25π/18 - 4π = -47π/18. That also seems correct. Yeah, I think I did it right.