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  • Bisectors of interior ∠B and exterior ∠ACD of a ∆ ABC intersect at the point T. Prove that ∠ BTC = ∠ BAC.

Bisectors of interior \angle B and exterior \angle ACD of a \triangle ABC intersect at the point T. Prove that \angle BTC = \frac{1}{2} \angle BAC.

Answers (1)

According to the question,
Bisectors of interior \angle B and exterior \angle ACD of a \triangle ABC intersect at the point T
\angle TBC =\frac{1}{2} \angle ABC                          … (1)
And \angle TCD =\frac{1}{2} \angle ACD                  … (2)

Now from \triangle ABC we have
\angle BAC + \angle ABC = \angle ACD … (3) (exterior angle is equal to the sum of interior opposite angles)
And from \triangle TBC we have
\angle BTC + \angle TBC = \angle TCD (exterior angle is equal to the sum of interior opposite angles)
Or \angle BTC + \frac{1}{2} \angle ABC =\frac{1}{2} \angle ACD using (1 and 2)
Or \angle BTC = \frac{1}{2}(\angle ACD - \angle ABC)
Using (3), \angle ACD - \angle ABC = \angle BAC
So,

\angle BTC = \frac{1}{2} \angle BAC

Hence Proved

Posted by

infoexpert26

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