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.
Details: Click on the figure below.
Friday, March 29, 2019
Geometry Problem 1429: Triangle, Median, 15-30 degree, Angle, Congruence
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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