Question

bisector of interior angleB and exterior angle ACD of a triangle aABC interesct at the point T. Prove that angleBTC=¹/² angle BAC

Answer 1

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Answer 2

Answer:

Since,BT bisects ∠ABC, then.,

∠TBC =∠ABT=1/2 ∠ABC

Since, TC bisects ∠ACX, then,

∠TCX=∠ACT=1/2 ∠ACX

Since, ∠TCX is an exterior angle of ΔTBC

∠TCX = ∠TBC+∠BTC [ exterior angle theorem]......{i}

since,∠ACX is an exterior angle of ΔABC

∠ACX= ∠BAC+∠ABC [exterior angle theorem]

==>1/2∠ACX=1/2 ∠BAC +1/2 ∠ABC

==>∠TCX= 1/2 ∠BAC + ∠TBC....{ii}

From {i} and {ii}...

∠TBC + ∠BTC= 1/2 ∠BAC + ∠TBC

==> ∠BTC = 1/2 ∠BAC.

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