Question

I fig., $AC>AB$ and AD is the bisector of $\angle A$. show that $\angle ADC>\angle ADB$

Answer 1

In $\Delta ABC$ ,

$AC>AB$ (given)

So,

$\angle B>\angle C$ ............... (1) ( Angle opposite to the larger side is Larger )

In $\Delta ABD$,

$\angle 1+\angle B+\angle ADB={180}^{\circ }$

$\angle B={180}^{\circ }-\angle 1-\angle ADB$ .................... (2)

In $\Delta ADC$,

$\angle 2+\angle C+\angle ADC={180}^{\circ }$

$\angle C={180}^{\circ }-\angle 2-\angle ADC$ .................... (3)

From (1), (2) and (3)

${180}^{\circ }-\angle 1-\angle ADB$$>{180}^{\circ }-\angle 2-\angle ADC$

${180}^{\circ }-\angle 1-\angle ADB$$>{180}^{\circ }-\angle 1-\angle ADC$ ( $\angle 1=\angle 2$ )

$-\angle ADB>-\angle ADC$

$\Rightarrow \angle ADB<\angle ADC$