# 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.

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)