Question
In the figure, given below, $A B=A C$.Prove that: $\angle B O C=\angle A C D$.
✓
Answer
Let $\angle A B O=\angle O B C=x$ and $\angle A C O=\angle O C B=y$ In $\triangle \mathrm{ABC}$
$\angle B A C=180^{\circ}-2 x-2 y\ldots. (i)$
Since, $\angle B=\angle C\ldots[\mathrm{AB}=\mathrm{AC}]$
$\frac{1}{2} \mathrm{\sim B}=\frac{1}{2} \mathrm{C}$
$ \Rightarrow \mathrm{x}=\mathrm{y}$
Now,
$\angle \mathrm{ACD}=2 \mathrm{x}+\angle \mathrm{BAC} \ldots[$ Exterior angle is equal to sum of opp. interior angles $]$
$\angle \mathrm{ACD}=2 \mathrm{x}+180^{\circ}-2 \mathrm{x}-2 \mathrm{y} \ldots[$ From $(i) ]$
$\angle \mathrm{ACD}=180^{\circ}-2 \mathrm{y} \dots....(i)$
In $\triangle O B C_1$
$ \angle B O C=180^{\circ}-x-y$
$\Rightarrow \angle B O C=180^{\circ}-\mathrm{y}-\mathrm{y} \ldots[$ Already proved $]$
$\Rightarrow \angle \mathrm{BOC}=180^{\circ}-2 \mathrm{y} ...(ii)$
From $(i)$ and $(ii)$
$\angle B O C=\angle A C D$
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