If $ \triangle A B C $ is isosceles with $ A B=A C $ and $ C(O, r) $ is the incircle of the $ \Delta A B C $ touching $ B C $ at $ L $, prove that $ L $ bisects $ B C $.

Given:

\( \triangle A B C \) is isosceles with \( A B=A C \) and \( C(O, r) \) is the incircle of the \( \Delta A B C \) touching \( B C \) at \( L \).

To do:

We have to prove that \( L \) bisects \( B C \).

Solution:

$AM$ and $AN$ are the tangents to the circle from $A$.

$AM = AN$

$AB = AC$        (given)

$AB - AN = AC - AM$

$BN = CM$

$BL$ and $BN$ are the tangents from $B$

$BL = BN$

Similarly,

$CL$ and $CM$ are tangents from $C$

$CL = CM$

Therefore,

$CL=BL$     (since $BN = CM$ and $BN=BL$)

Hence, $L$ bisects $BC$.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

28 Views

Kickstart Your Career

Get certified by completing the course

Get Started