Question

In triangle $ABC$, we have $AB=7,AC=8,BC=9$. Point $D$ is on the circumscribed circle of the triangle so that $AD$ bisects angle $BAC$. What is the value of $\frac{AD}{CD}$?

• A. $\frac{4}{3}$
• B. $\frac{3}{4}$
• C. $\frac{3}{5}$
• D. $\frac{5}{3}$

Answer 1

The correct option is D $\frac{5}{3}$

Let angle $BAC$ be $\theta$ , so angle $A$ is $2\theta$

Let angle $B$ be $\mu$

We have $cos2\theta =\frac{64+49-81}{112}=\frac{2}{7}$

$\Rightarrow cos\theta =\frac{3}{\sqrt{14}}$ and $sin\theta =\frac{\sqrt{5}}{\sqrt{14}}$

We have $cos\mu =\frac{64+81-49}{144}=\frac{2}{3}$

In triangle $ADC$ , by sine rule we get $\frac{AD}{sin(\theta +\mu )}=\frac{CD}{sin\theta }$

$\Rightarrow \frac{AD}{CD}=\frac{sin(\theta +\mu )}{sin\theta }=cos\mu +\frac{cos\theta sin\mu }{sin\theta }=\frac{2}{3}+1=\frac{5}{3}$