Question

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

Answer 1

Given In ∆ABC, produce SC to D and the bisectors of ∠ABC and ∠ACD meet at point T.

To prove `∠BTC = 1/2 ∠BAC`

Proof: In ∆ABC, ∠C is an exterior angle.

∴ ∠ACD = ∠ABC + ∠CAB  ...[Exterior angle of a triangle is equal to the sum of two opposite angles]

⇒ `1/2 ∠ACD = 1/2 ∠CAB + 1/2 ∠ABC`  ...[Dividing both sides by 2]

⇒ `∠TCD = 1/2 ∠CAB + 1/2 ∠ABC`  ...(i)  `[∵ "CT is a bisector of "∠ACD ⇒ 1/2 ∠ACD = ∠TCD]`

In ∆BTC, ∠TCD = ∠BTC + ∠CBT  ...[Exterior angle of a triangle is equal to the sum of two opposite interior angles]

⇒ `∠TCD = ∠BTC + 1/2 ∠ABC`   ...(ii) `[∵ "BT bisects of"  ∠ABC ⇒ ∠CBT = 1/2 ∠ABC]`

From equations (i) and (ii),

`1/2 ∠CAB + 1/2 ∠ABC = ∠BTC + 1/2 ∠ABC`

⇒ `∠BTC = 1/2 ∠CAB`

or `∠BTC = 1/2 ∠BAC`