Question

Pls answer Q. 48.

Need qualitative answer.
Need good explanation.

Answer 1

If bisectors of interior ∠B and exterior ∠ACD of ΔABC intersect at the point T , then ∠BTC =

(1) 2∠BAC

(2) ½∠BAC

(3) ½∠ABC

(4) 2∠ABC

Solution:

Given:

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

To Prove:
∠BTC = ½∠BAC

Proof:

∆ABC, ∠ACD is an exterior angle.

∴ ∠ACD = ∠ABC + ∠CAB

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

⇒½∠ACD =½∠CAB + ½∠ABC

[Dividing both sides by 2]

⇒ ∠TCD = ½∠CAB + ½∠ABC

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

In ∆, BTC

∴ ∠TCD = ∠BTC + ∠CBT

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

⇒ ∠TCD = ∠BTC + ½∠ABC . . . (ii)

[ ∵ bisector of∠ABC ⇒ ∠CBT =½∠ABC]

From equations (i) and (ii), we get

½∠CAB + ½∠ABC = ∠BTC + ½∠ABC

⇒ ½∠CAB = ∠BTC

or ½∠BAC = ∠BTC

Hence proved.

Answer 2

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