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bisector of interior angle b and exterior angle acd of a triangle abc intersect at point p if the d is the midpoint of the hypotenuse AC of a right triangle ABC proveq that BD=1/2AC

Posted by 〘』Ֆѧťᵞꭿ『〙 _ 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Seth 5 years, 1 month ago

Given:- ∆ABC is a triangle where BT and CT are internal and external bisectors of <ABC and <ACB respectively and they meet at T.

RTP:- <BTC = 1/2 <BAC

Construction:- Let, BC is produced to D and the external <ACD is formed.

Proof:-

In ∆ABC,

Exterior <ACD = interior opposite angle ( <ABC + <BAC )

=) 1/2 <ACD = 1/2 <ABC+ 1/2 <BAC

=) <TCD = <TBC + 1/2 <BAC ......( 1 )

In ∆TBC,

<TCD = <BTC + <TBC ........( 2 )

From ( 1 ) and ( 2 ),

<TBC + 1/2 <BAC = <BTC + <TBC

=) 1/2 <BAC = <BTC

=) <BTC = 1/2 <BAC

[ PROVED ]

1Thank You

ANSWER

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