Question

prove this............​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given :

BD = CD

AE is perpendicular to BC

To prove :

AB² + AC² = 2(AD² + BD²)

Proof :

By Pythagoras Theorem,

In triangle ABE ,

AB² = AE² + BE²

AB²= AE²+(BD-DE)²

AB²=AE² + BD² + DE² -2(BD)(DE) ----(1)

Similarly In triangle AEC ,

AC²= AE² + CE²

AC²= AE² +(BD+DE)² {BD=CD}

AC²= AE²+BD²+DE²+2(BD)(DE)-------(2)

In Triangle ADE,

AD²= AE²+DE² -----------------(3)

ADDING (1) & (2)

AB²+AC² = 2AE²+2BD²+2DE² –2(BD)(DE) +2(BD)(DE)

AB² + AC²= 2(AE²+DE²+BD²) {Since (3) }

AB²+AC²= 2(AD²+BD²)

$Hence \: Proved \:$