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°.

Very good simple solution.

DeleteFind E on DC such that < CBE = 3@

ReplyDeleteLet BD and BE extended meet the circle ABC at X and Y.

AX = CY so ACYX is an isoceles trapezoid.

Now < AEB = 7@ = < ABE

Hence AB = c = AE = CD

So AD = CE = b-c

Further < XAC = YCA

So tr.s AXD and CYE are congruent SAS.

< BDC = < BEA = 7@

Hence 7@ + 7@ + 4@ = 180 and therefore

@ = 10 and x = 40

Sumith Peiris

Moratuwa

Sri Lanka