Question

Bisector of interior angle B and exterior angle ACD of ∆ ABC intersect at a point T. Prove that angle BTC = 1/2 angle BAC
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Answer 1

Answer:

Step-by-step explanation:

in triangle ABC

∠BAC + ∠ABC =∠ ACD

DIVIDE BOTH SIDES BY 2

SO IT BECOMES

1/2 ∠BAC+ ∠TBC=∠TCD..........(1)

In triangle TBC

∠BTC +∠TBC =∠TCD..............(2)

on equating 1 and 2 we get

∠BTC +∠TBC = 1/2∠BAC + ∠TBC

cancel ∠TBC on both sides

∠BTC =1/2∠BAC

HENCE PROVED

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