Question
An isosceles $\triangle A B C$ has $A C=B C. C D$ bisects $A B$ at $D$ and $\angle C A B=55^{\circ}$.Find:$(i) \angle DCB;(ii) \angle CBD.$
✓
Answer
In $\triangle \mathrm{ABC}$,
$A C=B C\dots.......[$Given$]$
$\therefore \angle C A B=\angle C B D\dots........[$angles opp.to equal sides are equal$]$
$\Rightarrow \angle C B D=55^{\circ}$
In $\triangle \mathrm{ABC}$,
$ \angle \mathrm{CBA}+\angle \mathrm{CAB}+\angle \mathrm{ACB}=180^{\circ}$
but,$\angle \mathrm{CAB}=\angle \mathrm{CBA}=55^{\circ}$
$\Rightarrow 55^{\circ}+55^{\circ}+\angle \mathrm{ACB}=180^{\circ}$
$\Rightarrow \angle \mathrm{ACB}=180^{\circ}-110^{\circ}$
$\Rightarrow \angle \mathrm{ACB}=70^{\circ}$
Now,
In $\triangle A C D$ and $\triangle B C D$,
$ \mathrm{AC}=\mathrm{BC} \ldots . . .[$ Given$]$
$\mathrm{CD}=\mathrm{CD} \ldots . . . .[$ Common$]$
$\mathrm{AD}=\mathrm{BD} \ldots . . . .[$ Given: $\mathrm{CD}$ bisects $\mathrm{AB}]$
$\therefore \triangle \mathrm{ACD} \cong \triangle \mathrm{BCD}$
$\Rightarrow \angle \mathrm{DCA}=\angle \mathrm{DCB}$
$\Rightarrow \angle \mathrm{DCB}=\frac{\angle \mathrm{ACB}}{2}=\frac{70^{\circ}}{2}$
$\Rightarrow \angle \mathrm{DCB}=35^{\circ} .$
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