## Tuesday, March 11, 2014

### Geometry Problem 992: Triangle, Interior Point, Angles, 20, 30 , 40, 50 Degree

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view the complete problem 992

1. http://s25.postimg.org/7jzp7p7zz/Pro_992.png

Draw altitude CE of isosceles triangle ABC
We haves angles as shown on the sketch
D is the center of incircle of triangle AFC => angle DCF=10
So x= 180-60-10= 110

1. nice solution....

2. If E is the circumcenter of tr. ABD, then tr. ADE is equilateral (1) and, easily, tr. ACE and BEC are isosceles; with (1) CD is bisector of <ACE, making <BCD=30,hence <BDC=110.

Best regards

1. Why is tr. ADE equilateral, Stan?

2. Because <ABD=30 degs.

3. Would it be possible for you to sketch this out? I am having trouble visualizing and showing that ACE/BEC are isosceles and ACE is bisector. I can kind of see how that would happen but I don't know how to prove it

3. Excellent Stan!

4. Let AEBC be a kite with ABE equilateral and let AE meet BD at F

< FAB = FBA = 30 and so DA bisects < FAC
< AFD = FCD = 60
So DC must bisect < FAC
Hence < x = 180 - 40 -30 = 110

Sumith Peiris
Moratuwa
Sri Lanka

5. hi sumith, could you please provide a diagram? much appreciated.

6. t=20, 30+t=50, 2t=40, 90-3t=30, is the general form of the question. The answer is 150-2t=110.