Question

If $\Delta ABC$ is isosceles with AB = AC and C(O,r) is the incircle of the $\Delta ABC$ touching BC at L, prove that L bisects BC.

Answer 1

Given: ABC is an isosceles triangle.

C(O,r) is the incircle of $\Delta ABC$.

$\therefore$ O is the point of intersection of angle bisector. [The incenter of a triangle is the point of intersection of all the three interior angle bisectors of the triangle]

i,e. OB bisects $\angle B$ and OC bisects $\angle C.$

In triangle ABC, we have

$AB=AC$ (Given)

$\Rightarrow \angle C=\angle B$ (Since two sides are equal, angles opposite them also equal)

$\Rightarrow \angle OCL=\angle OBL$ (OB bisects $\angle B$ and OC bisects $\angle C$)

In $\Delta OCL$ and $\Delta OBL$, we have

$\angle OLB$ = $\angle OLC={90}^{\circ }$ [$\because$ Radius OL $\perp$ Tangent BC]]

$\angle OBL$ = $\angle OCL$ [Proved above]

$OL=OL$ [Common in both triangles]

$\therefore \Delta OCL\cong \Delta OBL$ [By AAS congruency rule]

$BL=CL$ [CPCT]

Thus, L bisects the side BC.