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.