Not too easy

Geometry Level 2

In triangle $ABC$ , angle $C$ is half of angle $B$ . $AD$ is the angle bisector of angle $BAC$ and $AB=CD$ , then $\angle BAC$ .

24 degree 120 degree 36 degree 80 degree 54 degree 72 degree 100 degree 154 degree

1 solution

Sanyam Goel
Feb 6, 2016

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

Not too easy

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