Question

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Answer 1

Answer:

In a triangle ABC, angle B is twice angle C. AD bisects angle BAC. AB = DC. Prove that angle BAC = 72 degrees?

In triangle ABC, we have

Angle B =2 Angle C

Conside Angle C =y then

Angle B =2y

AD is the bisector of Angle BAC

so, Let Angle BAD= Angle CAD = x

let BP be the bisector of Angle ABC . Join PD.

In triangle BPC,

Angle CBP= Angle BCP =y

BP=PC

Now, In triangle ABP and triangle DCP,

Angle ABP =Angle DCP =y

AB =DC already given

and BP=PC from above

so by SAS congruence creteria,

Triangle ABP congruence triangle DCP

Angle BAP =Angale CDP and AP=DP

Angle CDP=2x Angle ADP=DAP =x {Angle A =2x}

In triangle ABD,

Angle ADC =Angle ABD +Angle BAD

=>x+2x=2y+x

=> x=y

In trinagle ABC we have

Angles A+ B+C =180 degree

2x+2y+y=180 degree

=>5x=180 degree

=>x =36 degree

Hence, Angle BAC =2x= 72 degree

Step-by-step explanation:

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