Question
Find $x$ in Figure Given: $D A=D B=D C, B D$ bisects $\angle A B C$ and $\angle A D B=$$70^{\circ}$.
✓
Answer
In the figure,
$D A=D B=D C$
$B D$ bisects $\angle A B C$
and $\angle ADB =70^{\circ}$
But $\angle ADB +\angle DAB +\angle DBA =180^{\circ}$ (Angles of a triangle)
$ \Rightarrow 70^{\circ}+\angle DBA +\angle DBA =180^{\circ} \ldots \ldots . .(\because DA = DB )$
$\Rightarrow 70^{\circ}+2 \angle DBA =180^{\circ}$
$\Rightarrow 2 \angle DBA =180^{\circ}-70^{\circ}$
$=110^{\circ}$
$\therefore \angle DBA =\frac{110^{\circ}}{2}=55^{\circ} $
$\because B D$ is the bisector of $\angle A B C$,
$\therefore \angle DBA =\angle DBC =55^{\circ}$
But in $\triangle DBC$,
$D B=D C$
$\therefore \angle DCB =\angle DBC$
$\Rightarrow x=55^{\circ}$
$D A=D B=D C$
$B D$ bisects $\angle A B C$
and $\angle ADB =70^{\circ}$
But $\angle ADB +\angle DAB +\angle DBA =180^{\circ}$ (Angles of a triangle)
$ \Rightarrow 70^{\circ}+\angle DBA +\angle DBA =180^{\circ} \ldots \ldots . .(\because DA = DB )$
$\Rightarrow 70^{\circ}+2 \angle DBA =180^{\circ}$
$\Rightarrow 2 \angle DBA =180^{\circ}-70^{\circ}$
$=110^{\circ}$
$\therefore \angle DBA =\frac{110^{\circ}}{2}=55^{\circ} $
$\because B D$ is the bisector of $\angle A B C$,
$\therefore \angle DBA =\angle DBC =55^{\circ}$
But in $\triangle DBC$,
$D B=D C$
$\therefore \angle DCB =\angle DBC$
$\Rightarrow x=55^{\circ}$
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