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:

Please refer to both the attachments

Step-by-step explanation:

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noobiepro169@

Answer 2

$⚝ \: \: {\underline{\underline\mathfrak\red{Question:--}}}$

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

∠BTC = 1/2 ∠BAC

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$⚘ \: {\underline{\underline\mathfrak\red{Answer:--}}}$⠀

$❥ \: \: {\underline{\boxed{\bf\green{Given:}}}}$

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

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$❥ \: {\underline{\boxed{\bf\green{To \: \: prove:}}}}$

$\bf{∠BTC = \frac{1}{2} ∠BAC }$

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$❥ \: \: {\underline{\boxed{\bf\green{proof:}}}}$

$\bf{In \: Δ ABC, \: BC \: is \: extended \: to \: D}$

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❐ ∠ACD = ∠B + ∠A (Exterior ∠ prop.)

½ ∠ACD = ½∠B + ½∠A

∠1 = ½ ∠B + ½ ∠A ---------------(1)

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❐ In ΔTBC, BC is extended to D

∠1 = ∠3 + ∠T

∠1 = ½ ∠B + ∠T -----------------(2)

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❐ From (1) & (2)

½ ∠B + ∠T = ½ ∠B + ½ ∠A

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${\boxed{\bf\purple{∠BTC = \frac{1}{2} \: ∠BAC }}}$

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❒ Hence, proved.

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