Answer:
AC² + BD² = 2[AB² + BC²]
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.