Question

In triangle $ABC$, $a$, $b$, and $c$ are the sides and $A$, $B$, and $C$ are the respective opposite angles. $AD$ bisects angle $A$ and meets $BC$ at $D$. Then if $x=\overline{CD}$ and $y=\overline{BD}$ the correct proportion is :

• A. $x/c=a/(b+c)$
• B. $x/b=a/(a+c)$
• C. $y/c=c/(b+c)$
• D. $y/c=a/(b+c)$
• E. $x/y=c/b$

Answer 1

The correct option is D $y/c=a/(b+c)$

$a=x+y\frac{a}{\sin 2\theta }=\frac{b}{\sin B}=\frac{c}{\sin C}\frac{c}{\sin (C+\theta )}=\frac{y}{\sin \theta }=\frac{AD}{\sin B}\frac{b}{\sin (B+\theta )}=\frac{x}{\sin \theta }=\frac{AD}{\sin C}y=\frac{AD\sin \theta }{\sin B}x=\frac{AD\sin \theta }{\sin C}y\sin B=x\sin Cy\left(\frac{b\sin C}{c}\right)=x\sin Cyb=cxyb+cy=cx+cyy(b+c)=c(x+y)\therefore \frac{y}{c}=\frac{a}{b+c}$