Thursday, March 8, 2012

Problem 736: Triangle, Altitude, Perpendicular, Angles, Cyclic Quadrilateral

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 736 details.

Online Geometry Problem 736: Triangle, Altitude, Perpendicular, Angles, Cyclic Quadrilateral.

5 comments:

  1. http://img826.imageshack.us/img826/5130/problem736.png

    Quadrilateral DBEH is concyclic
    In right triangles AHB and BHC we have BD.BA=BH^2=BE.BC => quadrilateral ADEC is concyclic
    Let O is the center of cycle ADEC. (See sketch)
    In circle DBEH, ∠ (HBE)= ∠ (HDE)=33 => ∠ (EDC)=33-21=12
    Angles at the center of circle ADEC ∠ (AOE)=2 *57=114 and ∠ (EOC)=2*12=24
    So ∠ (AEB)=x=1/2*(114+24)=69

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  2. Triangles BDH and BHA are similar
    So are trangles BEH and BHC
    Therefore BD.BA = BH^2 = BE.BC
    So D,A,C,E are concyclic
    <BDE = <ECA = 57 deg
    <EDC = 90 - 57 -21 = 12 deg
    <EAC = <EDC = 12 deg
    Hence x = 57 deg + <EAC = 57 + 12 = 68 deg

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  3. The solution is uploaded to the following link:

    https://docs.google.com/open?id=0B6XXCq92fLJJeHNTbEVCdnk1LU0

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  4. BDHE is concyclic hence < BDE = < BHE = < BCH = 57 so ADEC is also concyclic

    Hence < CDE = 12 = < EAC. Therefore x =12 +57 = 69

    Sumith Peiris
    Moratuwa
    Sri Lanka

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