Question

In figure , $segAD\perp segBC$. $segAE$ is the bisector of $<CAB$ and $C-E-D$ . Prove that , $\angle DAE=\frac{1}{2}(\angle C-\angle B)$

Answer 1

As AE bisects $\angle A$, then,

$\angle BAE=\angle CAE$

In $\Delta ADC$,

$\angle ADC+\angle DAC+\angle ACD={180}^{\circ }$

${90}^{\circ }+\angle DAC+\angle C={180}^{\circ }$

$\angle C={90}^{\circ }-\angle DAC$

In $\Delta ADB$,

$\angle ADB+\angle DAB+\angle ABD={180}^{\circ }$

${90}^{\circ }+\angle DAB+\angle B={180}^{\circ }$

$\angle B={90}^{\circ }-\angle DAB$

$\angle C-\angle B=\angle DAB-\angle DAC$

$\angle C-\angle B=\left(\angle BAE+\angle DAE\right)-\left(\angle CAE-\angle DAE\right)$

$\angle C-\angle B=\angle BAE+\angle DAE-\angle BAE+\angle DAE$

$\angle C-\angle B=2\angle DAE$

$\frac{1}{2}\left(\angle C-\angle B\right)=\angle DAE$