Friday, March 29, 2019

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

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.

Geometry Problem 1429: Triangle, Median, 15-30 degree, Angle, Congruence.
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7 comments:

  1. UnknownMarch 31, 2019 at 8:07 AM

    1)Reflect triangleBDC across BD
    2)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

    ReplyDelete
  2. AnonymousMarch 31, 2019 at 5:50 PM

    Make Equilateral OBC, so <ABO=15 degree;
    connect 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.

    ReplyDelete
  3. Pradyumna AgasheMarch 31, 2019 at 10:55 PM

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

    ReplyDelete
  4. APOSTOLIS MANOLOUDISApril 12, 2019 at 3:18 AM

    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.

    ReplyDelete
  5. ANGLE BISECTORJune 2, 2019 at 7:03 PM

    Use the sine rule and its a cakewalk.

    ReplyDelete
  6. UnknownApril 2, 2020 at 10:30 AM

    Use congruancy rule

    ReplyDelete
  7. c.t.e.o.March 20, 2023 at 1:48 PM

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

    ReplyDelete

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