In the figure, ABC is a triangle in which \(\angle\) ABC > 90 and AD \(\perp\) CB produced. Prove that \({AC}^2\) = \({AB}^2\) + \({BC}^2\) + 2BC.BD.

Solution :

Given : ABC is triangle in which \(\angle\) ABC > 90 and AD \(\perp\) CB produced.triangle

To Prove : \({AC}^2\) = \({AB}^2\) + \({BC}^2\) + 2BC.BD

Proof : Since \(\triangle\) ADB is a right triangle, angled at D. Therefore, by Pythagoras theorem,

\({AB}^2\) = \({AD}^2\) + \({DB}^2\)           ……..(1)

Again, in triangle ADC is a right triangle, angled at D.

Therefore, by Pythagoras theorem, we have :

\({AC}^2\) = \({AD}^2\) + \({DC}^2\)

\(\implies\)  \({AC}^2\) = \({AD}^2\) + \({DB + BC}^2\)

\(\implies\)  \({AC}^2\) = \({AD}^2\) + \({DB}^2\) + \({BC}^2\) + 2BC.BD

\(\implies\)  \({AC}^2\) = \({AB}^2\) + \({BC}^2\) + 2BC.BD             (by using 1)

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