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
See also:
Complete Problem 481

Level: High School, SAT Prep, College geometry

3 comments:

  1. 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

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  2. I think its still easier...

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

    as dessired.

    ReplyDelete
  3. Let BG and AF meet at X

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

    Sumith Peiris
    Moratuwa
    Sri Lanka

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