Wednesday, May 4, 2016

Geometry Problem1210. Circle, Tangent Line, Secant, Chord, Collinear Points

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.


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Geometry Problem1210. Circle, Tangent Line, Secant, Chord, Collinear Points
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2 comments:

  1. Pradyumna AgasheMay 10, 2016 at 8:14 PM

    We can see that EC is bisector of Angle AEB, lets assume EC intersects AB at point J.
    By Ceva's theorem
    AH/HG=(AJ/BJ).(BP/PG)
    Lets Assume ED extended meets AG at point H', Consider triangle AJG and EDH' as transversal and applying Menelaus' theorem
    AH'/H'G=(AD/DJ)*(JE/GE) .............(1)

    If we can prove AH/HG=AH'/H'G, then H and H' coincide, and we get desired result that E,D,H are collinear.
    Conisder triangle AJE and Triangle EBG
    Angle AEJ= Angle BEG = x,
    if We assumne angle EBJ = y then Angle EBG = 180 - y
    Applying sine law we get JE/GE = AJ/BG
    We can write AH'/H'G=(AD/DJ)*(AJ/BG)................(2)

    Applying Menelaus' theorem for Triangle BDP with EC as transversal, we get
    BG/GP = (BJ/DJ)*(DC/PC)
    also if we join BC,it is bisectors of Angle DBP hence DC/PC = BD/BP = AD/BP
    We get BG/PG = (BJ/DJ)*(AD/BP) or BG*DJ =(PG/BP)*AD*BJ
    replacing BG*DJ in eq 2 , we have AH'/H'G=(AJ/BJ)*(BP/PG) = AH/HG
    Hence H' and H coincide and E,D,H are collinear.

    ReplyDelete
  2. Pradyumna AgasheMay 11, 2016 at 1:56 AM

    Correction: if We assumne angle EAJ = y then Angle EBG = 180 - y

    ReplyDelete

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