In figure below, $\angle B"

Given:

$\angle B<90^o$ and segment $AD \perp BC$.

To do:

We have to show that $b^2=h^2+a^2+x^2-2ax$.
Solution:
In $\triangle ADC$, by using Pythagoras theorem,

$ AC^2=AD^2+DC^2$

$b^2=h^2+(a-x)^2$

$b^2=h^2+a^2+x^2-2ax$

Hence proved.

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Updated on: 10-Oct-2022

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