Geometry Problem

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

Click the figure below to see the complete problem 671.

## Thursday, September 22, 2011

### Problem 671: Triangle, Cevian, Angles, Congruence

Labels:
angle,
cevian,
congruence,
triangle

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http://img838.imageshack.us/img838/3824/problem671.png

ReplyDeleteDraw triangle BDE as image of triangle BDA ( see picture)

Note that BE=CD and (DCB)=(EBC)=4. alpha

And DBCE is an isosceles trapezoid

(BDE)=(BDA)= 180-7.alpha

In triangle ABD , x+180-7.alpha+3.alpha= 180 >> x=4. alpha

In triangle ABC , 7.alpha+4.alpha+4.alpha=180 >> alpha=10 and x=40

Peter Tran

Okay, I have a really simple solution.

ReplyDeleteAt first the problem seems to be awful but it can be taken down faster.

Let P be a point on BC so that DB = DP, hence <CDP = 3a and then ∆CDP = ∆ABD and x = 4a, hence a = 10° whereat x = 40°.