Bisectors of interior angle B nd exterior angle ACD of a ∆ ABC interse...
Proof: Angle BTC = 1/2 Angle BAC
Given
ABC is a triangle with bisectors of interior angle B and exterior angle ACD intersecting at point T.
To Prove
Angle BTC = 1/2 Angle BAC
Proof
Let us consider the triangle ABC:
- Let angle BAC be denoted by x.
- Let angle ABC be denoted by y.
- Let angle BCA be denoted by z.
By the angle bisector theorem,
- Angle ABT = Angle TBC = y/2 (as BT is the angle bisector of angle B).
- Angle ATC = (180 - z)/2 = 90 - z/2 (as TC is the angle bisector of angle B).
Also, angle ACD = x + y (as angle ACD is the exterior angle of angle B).
Hence, angle BTC = Angle ABT + Angle ATC = y/2 + 90 - z/2.
Now, let us look at triangle ACD:
- Angle ADC = x + y (as it is the exterior angle of angle B).
- Angle DAC = 180 - z (as the angles of triangle ADC add up to 180).
By the exterior angle theorem, angle ACT = Angle ADC + Angle DAC = x + y + 180 - z.
But, we know that angle ATC = 90 - z/2. Hence, angle ACT = 2(90 - z/2) + x + y - 180 = x + y - z.
Therefore, angle BTC = y/2 + 90 - z/2 = (y + x)/2 = 1/2 Angle BAC.
Conclusion
Thus, we have proved that angle BTC = 1/2 Angle BAC.