Geometry Problem

Click the figure below to see the complete problem 495 about Triangle, 120 Degrees, Angle Bisectors.

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Complete Problem 495

Level: High School, SAT Prep, College geometry

## Sunday, August 8, 2010

### Problem 495: Triangle, 120 Degrees, Angle Bisectors

Labels:
120,
angle bisector,
degree,
triangle

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From Tr. DAC Ang.BDF = (60 + C)/2 = 30 + C/2 = x + C/2 from Tr. DFC. Hence x = 30 deg.

ReplyDeleteVihaan, Dubai

To Vihaan

ReplyDeleteYou mention that “From Tr. DAC Ang.BDF = (60 + C)/2 “ . I am not sure how do you get this

I understand that DF is not angle bisector of angle BDA per problem statement.

Peter Tran

BF is the external bisector of Ang. DAC and F is the intersection of BF & CF which is internal bisector of ang. C. Hence DF is the external bisector of ang. ADC.

ReplyDeleteWould that stand to reason?

Vihaan

Thanks Vihaan

ReplyDeletePeter Tran

Angle Bisector of <DAC hits FC at point P which is the incenter of triangle DAC. If <FCA is a, then <AFC is 60-a. <ADP=<CDP=(180-(60+2a))/2=60-a making FDAP an circumscribed quadrilateral. Therefore <CFD=<DAP=30

ReplyDeleteIvan Bazarov

The result follows upon observing that FD is indeed the external bisector in Tr. ADC

ReplyDelete