Question
An isosceles $\triangle ABC$ has $AC = BC. CD$ bisects $AB$ at $D$ and $\angle CAB = 55^o$.Find:$(i)\angle DCB(ii)\angle CBD.$
✓
Answer
In $\triangle \mathrm{ABC}$,
$A C=B C.......[$Given$]$
$\therefore \angle C A B=\angle C B D........[$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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