Question

In an isosceles triangle, $$ABC$$ with $$AB=AC$$ the bisectors of $$\mathrm{\angle }B$$and $$\mathrm{\angle }C$$ intersect at each other at $O$. Join $$A$$ to $$O$$. Show that:
i) $$OB=OC$$
ii) $$AO$$ bisects $\angle A$

Answer 1

Step 1 : Prove that $$OB=OC$$.

Given,

In isosceles $$\mathrm{\Delta }ABC$$, we have
$$AB=AC$$
$$OB$$ is the bisector of $$\mathrm{\angle }B$$
So, $$\mathrm{\angle }ABO=\mathrm{\angle }OBC={\displaystyle \frac{1}{2}}\mathrm{\angle }B$$
$$OC$$is the bisector of $$\mathrm{\angle }C$$
So,$$\mathrm{\angle }ACO=\mathrm{\angle }OCB={\displaystyle \frac{1}{2}}\mathrm{\angle }C$$
Now,
$\Rightarrow$ $$AB=AC$$ [ given ]
$\Rightarrow$ $$\mathrm{\angle }ACB=\mathrm{\angle }ABC$$ [ angles opposite to equal sides are equal]
$\Rightarrow$$${\displaystyle \frac{1}{2}}\mathrm{\angle }ACB={\displaystyle \frac{1}{2}}\mathrm{\angle }ABC$$

$\Rightarrow$ $\frac{1}{2}\mathrm{\angle }B=\frac{1}{2}\mathrm{\angle }\mathrm{C}$ [ by equation (1) and equation (2) ]
$\Rightarrow$ $$\mathrm{\angle }OCB=\mathrm{\angle }OBC$$
$\Rightarrow$ $\therefore OB=OC$ [ sides opposite to equal angles are equal ]

Hence proved, $\therefore OB=OC$.

Step 2 : Prove that $$AO$$ bisects $$\mathrm{\angle }A$$.

Proof:
We already proved that $$OB=OC$$
in $$\mathrm{\Delta }ABC$$ and $$\mathrm{\Delta }ACO$$ we have,
$\Rightarrow$ $$AB=AC$$ [given]
$\Rightarrow$ $$AO=AO$$ [common]
$\Rightarrow$ $$OB=OC$$ [ proven equal ]
$\Rightarrow$ $\therefore \mathrm{\Delta }ABO\cong \mathrm{\Delta }ACO$ [ Side-Angle-Side congruence rule]
$\Rightarrow$$\therefore \mathrm{\angle }OAB=\mathrm{\angle }OAC$ [ since corresponding parts of congruent triangles are equal]
That means, $$AO$$ bisects $$\mathrm{\angle }A$$

Hence proved, $$AO$$ bisects $$\mathrm{\angle }A$$.