Geometry Problem. Post your solution in the comments box below.

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

Click the figure below to view the complete problem 992

## Tuesday, March 11, 2014

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

Labels:
20,
30 degrees,
40 degrees,
50,
angle,
triangle

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

ReplyDeleteDraw 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

nice solution....

DeleteIf 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.

ReplyDeleteBest regards

Why is tr. ADE equilateral, Stan?

DeleteBecause <ABD=30 degs.

DeleteWould 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

DeleteExcellent Stan!

ReplyDeleteLet AEBC be a kite with ABE equilateral and let AE meet BD at F

ReplyDelete< 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

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

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

ReplyDelete