Question

In a right angle triangle, right angled at A,AD perpendicular to BC. Prove AB^2+CD^2=AC^2+BD^2

Answer 1

In triangle ABD, From Pythagoras Theorem,

${ab}^{2} = {bd}^{2} + {ad}^{2} \\ {ab}^{2} - {bd}^{2} = {ad}^{2} \: \: (1)$
In triangle ACD,

${ac}^{2} = {ad}^{2} + {cd}^{2} \\ {ac}^{2} - {cd}^{2} = {ad}^{2} \: \: (2)$

From (1) and (2),

${ab}^{2} - {bd}^{2} = {ac}^{2} - {cd}^{2} \\ {ab}^{2} + {cd}^{2} = {ac}^{2} + {bd}^{2}$