Monday, July 26, 2010

Problem 481: Triangle, Altitude, Cevian, Perpendicular, Angle

Geometry Problem
Click the figure below to see the complete problem 481 about Problem 481: Triangle, Altitude, Cevian, Perpendicular, Angle.

Problem 481: Triangle, Altitude, Cevian, Perpendicular, Angle
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Complete Problem 481

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 3:25 PM
Labels: , , , ,

4 comments:

  1. Peter TranJuly 27, 2010 at 10:30 AM

    Let (XYZ) denote angle XYZ
    Note that quadrilateral ABFE is cyclic . (AEB)=(AFB)=90
    We have (BAF)=(BEF) ( both angle face the same arc BF)
    Angle alpha = (ABF)- (EBF) and angle beta= (EBG)-(EBF)
    But ( ABF)= (EBG) ( both angle complementary to (BAF) and ( BEF) )
    So alpha=beta

    Peter Tran

    ReplyDelete
  2. AnonymousJuly 27, 2010 at 8:32 PM

    I think its still easier...

    AEFB cyclic implies that <ABE = <AFE=90-<BFG=90-(90-<FBG)=<FBG

    as dessired.

    ReplyDelete
  3. Sumith PeirisJune 21, 2016 at 2:48 AM

    Let BG and AF meet at X

    Then < ABE = < AFE = < XFG = < FBX

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Ignacio Larrosa CañestroApril 2, 2017 at 9:52 AM

    ABFE is cyclic ==> <AFE = <FBG and we conclude that <ABE = <CBG.

    ReplyDelete

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