Question
In an isosceles triangle $ABC$ , with $AB = AC$, the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$ . Join A to $O$ . Show that $O B=O C$ and $A O$ bisects $\angle A$.
✓
Answer
Given: In $\triangle A B C, A B=A C$, the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$. Construction: Joint $A$ to $O$
To prove: $O B=O C$ and $A O$ bisects $A$.
Proof: $A B=A C \ldots$ [Given]
$\therefore \angle B=\angle C \ldots[\angle$ s opposite to equal side of a $\triangle]$
$\therefore \frac{1}{2} \angle B=\frac{1}{2} \angle C$
$\therefore \angle OBC =\angle OCB \ldots$ [As BO bisects $\angle B$ and $CO$ bisects $\angle C ]$
$\therefore O B=O C \ldots$ [Sides opposite to equal $\angle$ s of a $\triangle$ ]
In $\triangle O A B$ and $\triangle O A C$,
$A B=A C \ldots$ [Given]
$O B=O C \ldots$ [As proved above]
$OA = OA \ldots$ [Common]
$\therefore \triangle OAB \cong \triangle OAC . . .$ [By $SSS$ property]
$\therefore \angle OAB = \angle OAC . . . $[c.p.c.t.]
$\therefore AO$ bisects $\angle A$
To prove: $O B=O C$ and $A O$ bisects $A$.
Proof: $A B=A C \ldots$ [Given]
$\therefore \angle B=\angle C \ldots[\angle$ s opposite to equal side of a $\triangle]$
$\therefore \frac{1}{2} \angle B=\frac{1}{2} \angle C$
$\therefore \angle OBC =\angle OCB \ldots$ [As BO bisects $\angle B$ and $CO$ bisects $\angle C ]$
$\therefore O B=O C \ldots$ [Sides opposite to equal $\angle$ s of a $\triangle$ ]
In $\triangle O A B$ and $\triangle O A C$,
$A B=A C \ldots$ [Given]
$O B=O C \ldots$ [As proved above]
$OA = OA \ldots$ [Common]
$\therefore \triangle OAB \cong \triangle OAC . . .$ [By $SSS$ property]
$\therefore \angle OAB = \angle OAC . . . $[c.p.c.t.]
$\therefore AO$ bisects $\angle A$
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