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

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

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## Friday, March 29, 2019

### Geometry Problem 1429: Triangle, Median, 15-30 degree, Angle, Congruence

Labels:
15 degree,
30 degrees,
angle,
congruence,
geometry problem,
median,
triangle

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1)Reflect triangleBDC across BD

ReplyDelete2)get triangleBDT ≅BDC

So,DT=DC

=DA

3)get Equilateral triangleBCT

so,BC=CT=TB

4)triangleATC is circumed by circle which center is D

So,angleATC=90degree

5)From angleABT=15degree

angleBTC=60degree

angleATC=90degree

So,angleTAB=15

Get AT=TB

6)triangleATC

Get angleTAC=TCA=45degree

7)AngleBAC=TAC-TAB

=45-15

=30

Make Equilateral OBC, so <ABO=15 degree;

ReplyDeleteconnect OA;

DCBO is kite. So DC=OD=AD, so <AOC =90 degree;

<ABO=15 degree, so <BAO =180- 90-15 = 15 degree;

so OA=OB=OC; O is center Circle for ACB; so <BAC =1/2 COB =30 degree.

Draw an equilateral triangle BCE such that Angle ABE=15 deg, Triangle BDC is congruent

ReplyDeleteto Triangle BDE by SAS, hence DE=DC. Since AD=DE=DC, Angle AEC=90 deg, we get Angle AEB=150 Deg.

Hence Triangle ABE is isosceles triangle with AE=BE.

We have AE=BE=CE, hence E must be circumcenter of Tr.ABC. since Angle BEC=60 Deg,

Angle BAC=60/2=30 Deg.

Let ∆ AED=equilateral (EC∩BD=F not extension) => ∆ DFC∼ ∆ DBC => AD^2=DC^2=DF*DB=> ∆ ADF∼ ∆ ABD => AE=ED=EF=> <DAB=15+15=30.

ReplyDeleteUse the sine rule and its a cakewalk.

ReplyDeleteUse congruancy rule

ReplyDeletehttps://photos.app.goo.gl/6457rBjdMRE2LZA58

ReplyDelete