Question

In $\Delta$ABC, AB = AC and the bisectors of angles B and C intersect at point O. Prove that BO = CO and the ray AO is the bisector of angle BAC.
[4 MARKS]

Answer 1

Concept : 1 Mark
Process : 2 Marks
Proof : 1 Mark

In $\Delta$ ABC , we have
AB = AC
$\Rightarrow \angle$B = $\angle$C [Angles opposite to equal sides are equal]
​$\frac{1}{2}$ $\angle$B = $\frac{1}{2}$ $\angle$C [OB and OC are bisectors of $\angle$B and $\angle$C]
$\Rightarrow$ $\angle$OBC = $\angle$OCB ….. (i)

$\Rightarrow$ OB = OC ……(ii) [Sides opp. To equal $\angle$s are equal]

Consequently, $\angle$ABO = $\angle$ACO ......(iii)


Now, in $\Delta$ABO and $\Delta$ACO, we have
AB = AC [Given]
$\angle$ABO = $\angle$ACO [From (iii)]
OB = OC [From (ii)]

$\therefore \Delta$ABO $\cong$ $\Delta$ACO [SAS criterion of congruence]

$\Rightarrow$ $\angle$BAO = $\angle$CAO [ C.P.C.T]

$\Rightarrow$ AO is the bisector of $\angle$BAC