Bisectors of interior ∠B and exterior ∠ACD of a ∆ ABC intersect at the point T. Prove that ∠BTC = 1/2 ∠BAC

Solution:

Given, ABC is a triangle

Bisectors of interior ∠B and exterior ∠ACD of the triangle intersect at the point T.

We have to prove that ∠BTC = 1/2 ∠BAC.

Extend BC to D.

Bisectors of ∠ABC and ∠ACD meet at T.

Considering triangle ABC,

∠C is the exterior angle

We know that the exterior angle is equal to the sum of the two opposite interior angles.

∠ACD = ∠ABC + ∠CAB

Dividing by 2 on both sides,

1/2 ∠ACD = 1/2 ∠ABC + 1/2 ∠CAB

Given, CT is the bisector of ∠ACD

∠ACD = ∠ACT + ∠TCD

∠ACT = ∠TCD

So, ∠ACD = ∠TCD + ∠TCD

∠ACD = 2 ∠TCD

∠TCD = 1/2 ∠ACD

Now, ∠TCD = 1/2 ∠ABC + 1/2 ∠CAB --------------------- (1)

Considering triangle BTC,

∠TCD = ∠BTC + ∠TBC
Given, BT is the bisector of ∠ABC

∠ABC = ∠ATB + ∠BTC

∠ATB = ∠BTC

∠ABC = ∠BTC + ∠BTC

∠ABC = 2∠BTC

∠BTC = 1/2 ∠ABC

Now, ∠TCD = ∠TBC + 1/2 ∠ABC ------------------------- (2)

From (1) and (2),

1/2 ∠ABC + 1/2 ∠CAB = ∠TBC + 1/2 ∠ABC

Canceling common term,

1/2 ∠CAB = ∠TBC

Therefore, ∠TBC = 1/2 ∠BAC

✦ Try This: In the above figure, l∥m. Find the values of x,y and z.

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 6

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NCERT Exemplar Class 9 Maths Exercise 6.4 Problem 2

Bisectors of interior ∠B and exterior ∠ACD of a ∆ ABC intersect at the point T. Prove that ∠BTC = 1/2 ∠BAC

Summary:

Bisectors of interior ∠B and exterior ∠ACD of a ∆ ABC intersect at the point T. It is proven that ∠BTC = 1/2 ∠BAC by exterior angle property which states that the exterior angles is always equal to the sum of the interior opposite angle

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