Sunday, August 8, 2010

Problem 495: Triangle, 120 Degrees, Angle Bisectors

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

Complete Problem 495

Level: High School, SAT Prep, College geometry

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

2. To Vihaan

You 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

3. 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.
Would that stand to reason?
Vihaan

4. Thanks Vihaan

Peter Tran

5. 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

Ivan Bazarov

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