Question

Bisectors of interior ∠B and exterior ∠ACD of a Δ ABC intersect at the point T.

Prove that ∠ BTC = ½ ∠ BAC.

Answer 1

Given: △ ABC, produce BC to D and the bisectors of ∠ABC and ∠ACD meet at point T.

To prove:

∠BTC = ½ ∠BAC

Proof:

In △ABC,∠ACD is an exterior angle.

We know that,

Exterior angle of a triangle is equal to the sum of two opposite angles,

Then,

∠ACD = ∠ABC + ∠CAB

Dividing L.H.S and R.H.S by 2,

⇒ ½ ∠ACD = ½ ∠CAB + ½ ∠ABC

⇒ ∠TCD = ½ ∠CAB + ½ ∠ABC …(1)

[∵CT is a bisector of ∠ACD⇒ ½ ∠ACD = ∠TCD]

We know that,

Exterior angle of a triangle is equal to the sum of two opposite angles,

Then in △ BTC,

∠TCD = ∠BTC +∠CBT

⇒ ∠TCD = ∠BTC + ½ ∠ABC …(2)

[∵BT is bisector of △ ABC ⇒∠CBT = ½ ∠ABC ]

From equation (1) and (2),

We get,

½ ∠CAB + ½ ∠ABC = ∠BTC + ½ ∠ABC

⇒ ½ ∠CAB = ∠BTC or ½ ∠BAC = ∠BTC

Hence, proved.