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