Question
✓
Answer
In $\triangle \text{APB}$ and $\triangle \text{APC}$
$\angle\text{APB}=\angle\text{APC}$ (Each)
AB = AC (Given)
AP = AP (Common)
$\therefore \triangle\text{APB}\cong\triangle \text{APC}$ (by RHS)
AB = AC (Sides opposite to equal angles of a triangle are equal)
$\Rightarrow \angle\text{B}=\angle\text{C}$ (By using CPCT).
$\angle\text{APB}=\angle\text{APC}$ (Each)
AB = AC (Given)
AP = AP (Common)
$\therefore \triangle\text{APB}\cong\triangle \text{APC}$ (by RHS)
AB = AC (Sides opposite to equal angles of a triangle are equal)
$\Rightarrow \angle\text{B}=\angle\text{C}$ (By using CPCT).
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