Question

If in a $△ABC,CD$ is the angular bisector of the $\angle ACB$, then $CD$ is equal to

• A. $\left(\frac{a+b}{2ab}\right)\cos \left(\frac{C}{2}\right)$
• B. $\left(\frac{a+b}{ab}\right)\cos \left(\frac{C}{2}\right)$
• C. $\left(\frac{2ab}{a+b}\right)\cos \left(\frac{C}{2}\right)$
• D. $\frac{b\sin A}{\sin (B+\frac{C}{2})}$

Answer 1

Answer: (C), (D)

The correct options are
C $\left(\frac{2ab}{a+b}\right)\cos \left(\frac{C}{2}\right)$
D $\frac{b\sin A}{\sin (B+\frac{C}{2})}$

$△CAB=△CAD+△CDB$
$\Rightarrow \frac{1}{2}ab\sin C=\frac{1}{2}.b.CD.\sin \frac{C}{2}+\frac{1}{2}.a.CD.\sin \frac{C}{2}$
$\Rightarrow CD=\frac{2ab}{a+b}\cos \frac{C}{2}$ (on evaluation)
In $△CAD,\frac{CD}{\sin A}=\frac{b}{\sin \angle CDA}$
or $CD=\frac{b\sin A}{\sin (B+\frac{C}{2})}$