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In triangle ABC sides AB and AC are produced to point P and Q such that angle PBC < angle QCB prove that AC >AB

Posted by Arushi Gaur 6 years, 4 months ago

CBSE > Class 09 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given : Side AB and AC of {tex}\triangle{/tex}ABC are extended to points P and Q respectively. Also ∠PBC < ∠QCB.
To Prove : AC > AB.
Proof : ∠PBC < ∠QCB
{tex}\therefore{/tex} -∠PBC > - ∠QCB
{tex}\therefore{/tex} 180^o - ∠PBC > 180^o - ∠QCB
{tex}\therefore{/tex} ∠ABC > ∠ACB
{tex}\therefore{/tex} AC > AB . . . [Side opposite to greater angle is longer]

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CBSE > Class 09 > Mathematics