Question

If the bisector of angle C of triangle ABC meets AB in D & the circumcircle in E then show that $\frac{CE}{DE}=\frac{{(a+b)}^2}{c^2}$

Answer 1

Applying sine law in $△BDE$
$\frac{\sin \frac{C}{2}}{DE}=\frac{\frac{\sin A}{ac}}{a+b}$ ...(1) $(\because BD=\frac{ac}{a+b})$
And Applying sine law in $△CBE$
$\frac{\sin B+\frac{C}{2}}{CE}=\frac{\sin A}{a}$ ...(2)
Divide (1) by (2)
$\frac{\sin \frac{C}{2}}{\sin \left(B+\frac{C}{2}\right)}.\frac{CE}{DE}=\frac{a+b}{ac}.a$
$\Rightarrow \frac{CE}{DE}=\frac{\frac{a+b}{c}.2\sin (B+\frac{C}{2})\cos \frac{C}{2}}{\sin C}$
$=\frac{a+b}{c}.\frac{\sin (B+C)+\sin B}{\sin C}=\frac{a+b}{c}.\frac{\sin A+\sin B}{\sin C}$
$\Rightarrow \frac{CE}{DE}=\frac{{(a+b)}^2}{c^2}$