Thursday, January 7, 2010

Problem 415: Right Triangle, Cevian, Angles, Congruence

Proposed Problem
Click the figure below to see the complete problem 415 about Right Triangle, Cevian, Angles, Congruence.

Problem 415: Right Triangle, Cevian, Angles, Congruence.
See also:
Complete Problem 415
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 6:40 AM
Labels: , , ,

5 comments:

  1. AnonymousJanuary 7, 2010 at 7:11 AM

    E on DC
    Let
    m(DBE)=2a
    m(BDE)=m(BED)=90-a
    m(ABE)=90-a
    BD=BE
    AB=AE
    {AD=BA-DE
    EC=DC-DE
    BA=DC}
    AD=EC
    x=2a
    4a=90
    x=45

    ReplyDelete
  2. UnknownJanuary 8, 2010 at 2:55 AM

    by sine rule on BAD triangle and BDC triangle

    ReplyDelete
  3. Ivan BazarovFebruary 22, 2014 at 5:01 PM

    E is on AC such that BD=BE. <ABD=90-3a so <BED=<BDE=90-a=<ABE making AB=AE and triangles ABE and CDB congruent leading to AE=CB. AB=AE=CB so x=45.

    ReplyDelete
  4. Sumith PeirisSeptember 28, 2015 at 10:52 AM

    Let BH be an altitude and let BE be the angle bisector of < HBC so that BH and BE trisect < 3@.

    Hence < ABH = x and < ABE = x + @ = < AEB so AB = AE = DC which in turn leads us to the conclusion that AD = EC

    Hence AB = BC and x = 45.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. MarcoDecember 9, 2023 at 11:16 PM

    [For easy typing, I use a instead of alpha]
    <ABD=90-3a
    <ADB=90+a
    x=90-2a

    In triangle ABD
    sin(90+a)/AB=sin2a/BD
    AB/BD=sin(90+a)/sin2a--------(1)

    In triangle BDC
    sinx/BD=sin3a/CD
    CD/BD=sin3a/sinx---------(2)

    Since AB=CD, (1)=(2)
    sin(90+a)/sin2a=sin3a/sinx
    cosa/2sinacosa=sin3a/sin(90-2a)
    cos2a=2sinasin3a
    cos2a=cos2a-cos4a
    cos4a=0
    4a=90
    2a=45
    x=90-2a=45

    ReplyDelete

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