In Fig. 6, if $AD \perp BC$, then prove that $AB^2 + CD^2 = BD^2 + AC^2$.

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Question

In Fig. 6, if $AD \perp BC$, then prove that $AB^2 + CD^2 = BD^2 + AC^2$.

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Accepted Answer

Step 1: In right $	riangle ABD$: $AB^2 = BD^2 + AD^2$ ... (i)
Step 2: In right $	riangle ADC$: $CD^2 = AC^2 - AD^2$ ... (ii)
Step 3: Adding (i) and (ii): $AB^2 + CD^2 = BD^2 + AD^2 + AC^2 - AD^2 = BD^2 + AC^2$

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