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
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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