Question

bisectors of interior Angle B and exterior angle ACD triangle ABC intersect at the point T prove that

∠BTC = 1/2 ∠BAC​

Answer 1

Answer:

$\huge\red{Answer}$

According to the problem

∠TBC=

2

1

∠B

and

∠TCD=

2

1

∠ACD

Now from Δ ABC we have

∠A+ ∠B= ∠ACD

And from ΔTBC we have

∠T+ ∠TBC= ∠TCD

or,

∠T+

2

1

∠B=

2

1

∠ACD

[Using (1) and (2)]

or,

∠T=

2

1

(∠ACD−

∠B)

or,

∠T=

2

1

∠A

.[ Using (3)]

or,

∠BTC=

2

1

∠BAC

Answer 2

Answer:

$\huge\red{Answer}$

According to the problem

∠TBC=

2

1

∠B

and

∠TCD=

2

1

∠ACD

Now from Δ ABC we have

∠A+ ∠B= ∠ACD

And from ΔTBC we have

∠T+ ∠TBC= ∠TCD

or,

∠T+

2

1

∠B=

2

1

∠ACD

[Using (1) and (2)]

or,

∠T=

2

1

(∠ACD−

∠B)

or,

∠T=

2

1

∠A

.[ Using (3)]

or,

∠BTC=

2

1

∠BAC