Question

Let $AD$ with $D$ on $BC$ be the bisector of $\angle A$ in $\Delta ABC$. If $b=AC,c=AB,m=CD$, and $n=BD$, then

• A. $\frac{b}{c}=\frac{2m}{n}$
• B. $\frac{b}{c}=\frac{m}{n}$
• C. $\frac{b}{c}=\frac{n}{m}$
• D. $\frac{b}{c}=\frac{m^2}{n^2}$

Answer 1

Answer: (B)

$\frac{b}{c}=\frac{m}{n}$

Given that $AC=b,AB=c,CD=m,BD=n$ and $AD$ is angular bisector of $A$

In triangle $ADC$, by sine rule we get $\frac{DC}{\sin \left(\angle DAC\right)}=\frac{AC}{\sin \left(\angle ADC\right)}$

$\Rightarrow \frac{DC}{AC}=\frac{\sin \left(\angle DAC\right)}{\sin \left(\angle ADC\right)}$

$\Rightarrow \frac{m}{b}=\frac{\sin \left(\angle \frac{A}{2}\right)}{\sin \left(\angle ADC\right)}$

In triangle $ADB$, By sine rule we get $\frac{DB}{\sin \left(\angle DAB\right)}=\frac{AB}{\sin \left(\angle ADB\right)}$

$\Rightarrow \frac{DB}{AB}=\frac{\sin \left(\angle DAB\right)}{\sin \left(\angle ADB\right)}$

$\Rightarrow \frac{n}{c}=\frac{\sin \left(\angle \frac{A}{2}\right)}{\sin \left(\angle ADB\right)}$

Observe that $\angle ADC+\angle ADB={180}^0$

$\Rightarrow \sin \left(\angle ADC\right)=\sin ({180}^0-\angle ADB)=\sin \left(\angle ADB\right)$

Therefore by equating above two equations, we get $\frac{m}{b}=\frac{n}{c}$

$\Rightarrow \frac{b}{c}=\frac{m}{n}$