Question

If in triangle ABC, AD is perpendicular to BC, let us prove that, AB^2+CD^2=AC^2+BD^2

Answer 1

Step-by-step explanation:

In ∆ACD and ∆ABD both are right angle triangle.

In ∆ACD

AC²=AD²+DC² ----------(1)

In ∆ABD

AB²=AD²+BD² ---------(2)

subtracting equation 1 & 2, we get,

AC²-AB²=AD²+DC²-AD²-BD²

AC²-AB²=DC²-BD²

AC²+BD²=AB²+DC²

Hence,

[AB²+CD²=AC²+BD²]