In a parallelogram ABCD, prove that (AC)2 + (BD)2= 2[(AB)? +(BC)?].​

AC² + BD² = 2[AB² + BC²]

Step-by-step explanation:

Let the parallelogram be ABCD with sides AB, BC, CD and AD. It also has diagonals AC and BD.

Since the diagonals are perpendicular and bisect each other at their mid-point, and P is the point of intersection of the diagonals, we have that AP = AC/2, PC = AC/2, PB = BD/2 and PD = BD/2.

Since APB forms a right angled triangles with length of sides AP, PB and AB where AB is the hypotenuse side, using Pythagoras' theorem, we have

AB² = AP² + PB²

Since AP = AC/2 and PB = BD/2, we have

AB² = (AC/2)² + (BD/2)²

AB² = AC²/4 + BD²/4   (1)

Also, BPC forms a right angled triangles with length of sides BP, PC and BC where BC is the hypotenuse side, using Pythagoras' theorem, we have

BC² = BP² + PC²

Since PC = AC/2 and PB = BD/2, we have

BC² = (AC/2)² + (BD/2)²

BC² = AC²/4 + BD²/4   (2)

Adding equations (1) and (2), we have

AB² = AC²/4 + BD²/4   (1)

+

BC² = AC²/4 + BD²/4   (2)

AB² + BC² = AC²/4 + BD²/4 +  AC²/4 + BD²/4

AB² + BC² = AC²/2 + BD²/2

Multiplying through by 2, we have

2[AB² + BC²] = AC² + BD²

So, AC² + BD² = 2[AB² + BC²]  which proves our expression.

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