Question

∆ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC.

Answer 1

∆ABC is a right triangle such that AB = AC. CD is the bisector of ∠C which intersects AB at D.

Let AB = AC = x units

In right ∆ABC,

${\mathrm{BC}}^2={\mathrm{AB}}^2+{\mathrm{AC}}^2$ (Pythagoras theorem)

$\Rightarrow {\mathrm{BC}}^2=x^2+x^2=2x^2$

$\Rightarrow \mathrm{BC}=\sqrt{2}x$

In ∆ABC, CD is the bisector of ∠C.

$\therefore \frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AC}}{\mathrm{BC}}$ (The bisector of an angle of a triangle divides the opposite side in the ratio of sides containing the angle)

$\Rightarrow \frac{\mathrm{AD}}{x-\mathrm{AD}}=\frac{x}{\sqrt{2}x}$

$\Rightarrow \sqrt{2}\mathrm{AD}=x-\mathrm{AD}$

$\Rightarrow \mathrm{AD}\left(\sqrt{2}+1\right)=x$

$\Rightarrow \mathrm{AD}=\frac{x}{\sqrt{2}+1}=x\left(\sqrt{2}-1\right)$

Now,

$\mathrm{AC}+\mathrm{AD}=x+\sqrt{2}\left(x-1\right)=\sqrt{2}x=\mathrm{BC}$

Hence proved.