 Hi and welcome to the session. Let's work out the following question. The question says prove that in a right angle's triangle the square of the hypotenuse is equal to sum of the squares of other two sides using the above result prove that the sum of squares on the sides of rhombus, sorry it's of the sides of rhombus is equal to sum of squares of its diagonals. Let's start with a solution to this question. We are given a right angle triangle ABC in which angle B is equal to 90 degree. We have to prove that AC square is equal to AB square plus BC square. Now we do some kind of construction here. We draw BD perpendicular to AC. Now let's start with the proof. We see that in triangle ADB and triangle ABC first thing angle A is equal to angle A because that is a common angle. Secondly angle ADB is equal to angle ABC. So angle ADB is equal to angle ABC. Each is 90 degree. Therefore we can say that triangle ADB is similar to triangle ABC by angle angle similarity criterion. Since these two triangles are similar therefore we can say that AD divided by AB is equal to AB divided by AC. They are the corresponding parts of similar triangles or we can say that AB square is equal to AD into AC. This we get by cross multiplication and we call this equation one. Now we compare two other triangles that is in triangle BDC and triangle ABC angle C is equal to angle C because that is a common angle. Secondly angle BDC is equal to angle ABC. Measure of each of them is 90 degree. Therefore we can say that triangle BDC is similar to triangle ABC again by angle angle similarity criterion. Since these two triangles are similar therefore we can say that DC divided by BC is equal to BC divided by AC because they are the corresponding parts of similar triangles. Now cross multiplying we say that BC square is equal to DC into AC and we call this equation two. Now adding one and two we get AB square plus BC square is equal to AD into AC plus BC into AC. Taking AC common from both the terms we have AC into AD plus CD. Now AD plus CD is again AC so this is equal to AC into AC and AC into AC is same as AC square. Therefore we have AB square plus BC square is equal to AC square. So this is what we were supposed to prove in first part of this question. Now let us start with the second part of this question that is the second part we are given a rhombus ABCD in which BD and AC are diagonals. So let this be the rhombus ABCD we see that BD and AC are the diagonals. Now what we are supposed to prove is that AB square plus BC square plus CD square plus AD square is equal to AC square plus BD square. So let us start with the proof for the second part. Since the diagonals of a rhombus they bisect each other at 90 degree therefore OD is equal to OB and OA is equal to OC. Now we see that in right angle triangle AOB AB square is equal to OA square plus OB square we call this 1 and this we get from the above theorem that we have just proved. Again in right angle triangle BOC we have BC square is equal to OC square plus OB square we call this 2. Now in triangle COD we have DC square is equal to OC square plus OD square and in triangle AOD we have AD square is equal to OD square plus OA square we call this equation 3 and this equation 4. Now adding all these 4 we get AB square plus BC square plus CD square plus AD square is equal to OA square plus OB square plus OB square plus OC square plus OD square plus OD square plus OA square that is equal to twice of OA square plus twice of OB square plus twice of OC square plus twice of OD square. But we see that OA is equal to OC is equal to half of AC and similarly OB is equal to OD square plus OB is equal to half of BD. Therefore we can say that AB square plus BC square plus DC square plus AD square is equal to twice of half of AC square plus half of BD square plus half of AC square plus half of BD square plus half of AC square plus half of BD square. This is equal to twice of 1 by 4 AC square plus 1 by 4 BD square plus 1 by 4 AC square plus 1 by 4 BD square plus 1 by 4 AC square plus 1 by 4 BD square. This is equal to twice of half of AC square plus half of BD square this is further equal to twice of half of AC square plus BD square 2 gets cancelled with 2 and we have AC square plus BD square AB square plus BC square plus DC square plus AD square is equal to AC square plus BD square that is this square plus this square plus this square plus this square is equal to sum of squares of the diagonals AC and BD. So this is what we were supposed to prove I hope that you understood the solution and enjoyed the session have a good day.