Thursday, November 25, 2010

Problem 543: Right Triangle, Altitude, Angle Bisector, Median

Geometry Problem
Click the figure below to see the complete problem 543 about Right Triangle, Altitude, Angle Bisector, Median.

Problem 543: Right Triangle, Altitude, Angle Bisector, Median.
Go to Complete Problem 543

Posted by Antonio Gutierrez at 1:46 PM
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3 comments:

  1. c .t . e. oNovember 27, 2010 at 12:55 PM

    DBE = 45° - C, FBE = 45° - C
    =>
    DBE = FBE

    ReplyDelete
  2. Sumith PeirisJuly 1, 2015 at 6:50 AM

    < ABD = < BCF = < CBF. But < ABE = < CBE. Hence < DBE = < FBE.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. MarcoDecember 24, 2023 at 10:12 PM

    Since BF is median & <BAC=90, we can assume ABC lies on a semi-circle with F as centre
    Let <BAC=<BAF=a, <ABD=90-a
    <ACB=90-a, since FB=FC (radii), <FBC=<FCB=<ACB=90-a
    So, <ABD=<FBC=90-a-------(1)
    <ABE=<EBC (given)--------(2)
    <EBD=<ABE-<ABD
    <EBF=<EBC-<FBC
    By (1) & (2), <EBD=<EBF & BE is the < bisector of <DBF

    ReplyDelete

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