In a triangle ABC, AD is drawn perpendicular to BC. How do I Prove that AB^2 - BD^2 = AC^2 - CD^2?

Question

I've been stuck on this for quite some time I just don't understand how to prove this equation. I usually know how to deal with these types of questions by looking at examples but it's not helping.

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Answer 1 · 2018-02-20T07:16:09

Please see below.

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In rt #DeltaADC#,

#rarrAD^2=AC^2-CD^2#.....[1]

In rt #DeltaADB#,

#rarrAD^2=AB^2-BD^2#.....[2]

From [1] and [2],

#AC^2-CD^2=AB^2-BD^2# Proved