In figure below, $\angle B"

Given:

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

To do:

We have to show that $b^2=a^2+c^2-2ax$.

Solution:

In $\triangle ADC$, by using Pythagoras theorem,

$ AC^2=AD^2+DC^2$

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

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

In $\triangle ADB$, by using Pythagoras theorem,

$ AB^2=AD^2+DB^2$

$c^2=h^2+x^2$.......(ii)

Substituting equation (ii) in (i), we get,

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

$b^2=c^2+a^2-2ax$

$b^2=a^2+c^2-2ax$

Hence proved.

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Updated on: 10-Oct-2022

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