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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 bisect BC.

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

To prove: L bisect BC

Proof: Construct the figure according to given condition.

AB = AC (given)

From the theorem which states that the lengths of two tangents drawn from external point to a circle are equal. ..... (1) 

As tangents AP and AQ are drawn from the external point A. 

AP = AQ 

Also, AB = AC

⇒ AP + PB = AQ + QC

⇒ AP + PB = AP + QC

⇒ PB = QC From (1) 

as tangents BP and BL are drawn from external point B, 

And tangents CQ and CL are drawn from external point C.

⇒ BP = BL ..... (3) 

CQ = CL ...... (4)

As we have proved PB 

= QC From 3 and 

4BL = CL

⇒ L bisects BC.

Hence proved.

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