Sunday, March 12, 2017

Geometry Problem 1322: Triangle, Angle Bisector, Circumcircle, Chord, Secant, Sum of two Angles

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Details: Click on the figure below..

Geometry Problem 1322: Triangle, Angle Bisector, Circumcircle, Chord, Secant, Sum of two Angles.

Posted by Antonio Gutierrez at 12:42 PM
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5 comments:

  1. c.t.e.oMarch 12, 2017 at 4:21 PM

    Draw circle O1 throw A,W,Y,B. Extend AX to P, WX to Q (P,Q on O1)
    Ang PWY = α, need to prove PWQ = β
    ZX//WP => <PWQ = β

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  2. APOSTOLIS MANOLOUDISMarch 12, 2017 at 10:32 PM

    Problem 1322
    let's say that <BAX=δ=<XZB(B,X,Z,A=concyclic) and <ΒWX=γ,but <XBZ=α=<YBW=<XAZ=<ZAW=<YAW.So the point A,B,Y and W are concyclic.Τhen
    <BWY=<BAY or γ+θ=δ+α (δ=β+γ in triangle ZXW) or θ=α+β.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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  3. PravinMarch 13, 2017 at 9:59 PM

    Draw WV // ZX (with V on BC).
    So <XWV = β (alternate angles).
    Enough to show <YWV = α.
    A,B,Y,W are concyclic (Note <WAY = <ZAX = <ZBX = α).
    So <YVW = <YXZ = <BAY = <BWY.
    Hence Triangles YVW, YWB are /// and <YWV = <WBY = α.

    ReplyDelete
  4. PravinApril 3, 2017 at 8:47 PM

    Easy to observe A,B,Y and W are concyclic.
    Denote <ABW = <AXZ = <AYW = γ
    Let AY and XW intersect at V.
    We evaluate <VYW in two ways:
    As an exterior angle of triangle VYW (i.e.) θ+γ
    As an exterior angle of triangle VXA (i.e.) (α+β)+γ
    Follows θ=α+β

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  5. Stan FULGERApril 5, 2017 at 2:18 AM

    Easy to see, ABXZ, ABYW are cyclic, hence <AZX=<AWY=180-<ABC ( 1 ). Since <AZX=<AWX+<ZXW+<ZAW, we get the required <XWY=<ZXW+<ZAW.

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