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# 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, (i) OB = OC (ii) AO bisects $$\angle{A}$$

In $$\triangle{ABC}$$, we have

AB = AC ...(Given)
Also, $$\angle{B}$$ = $$\angle{C}$$ ...(Since corresponding angles of equal sides are equal)

Multipling both sides by $$\frac{1}{2}$$ , we get,
($$\frac{1}{2}$$ ) $$\angle{B}$$ = ($$\frac{1}{2}$$ ) $$\angle{C}$$
$$\therefore$$ $$\angle{OBC}$$ = $$\angle{OCB}$$

Also, it is given that, OB and OC are bisectors of $$\angle{B}$$ and $$\angle{C}$$, respectively,

Therefore, $$\angle{OBA}$$ and $$\angle{OCA}$$
Therefore, OB = OC
...($$\because$$ corresponding sides of equal angles are equal)

Hence, part (i) is proved.

In $$\triangle{ABO}$$ and $$\triangle{ACD}$$, we have,

AB = AC ...(Given)
$$\angle{OBA}$$ = $$\angle{OCA}$$ ...(from part(i))
OB = OC ...(proved earlier)

Therefore, $$\triangle{ABO}$$ $$\displaystyle \cong$$ $$\triangle{ACO}$$ ...(By SAS congruency test)

Thus, $$\angle{BAO}$$ = $$\angle{CAO}$$ ...(By CPCT)

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