Friday, September 20, 2013

Geometry Problem 925: Cyclic Quadrilateral, Diagonal, Circle, Angle Bisector, Parallel

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

Click the figure below to see the complete problem 925.

Online Geometry Problem 925: Cyclic Quadrilateral, Diagonal, Circle, Angle Bisector, Parallel.

Posted by Antonio Gutierrez at 7:27 AM
Labels: , , , ,

4 comments:

  1. Peter TranSeptember 20, 2013 at 8:52 AM

    http://img39.imageshack.us/img39/8260/xhe.bmp

    Let EG cut CF at K. define angles x and y as shown on the sketch
    In triangle BEK we have y=∠ (BEG)- ∠ (CBD)=1/4(Arc(AB)+Arc(CD))- 1/2Arc(CD)= 1/4(Arc(AB)-Arc(CD))
    We have x= ½∠ (BFA)=1/2 x ½( Arc(AB)-Arc(CD))= 1/4(Arc(AB)-Arc(CD))
    So x=y and EG//FH

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  2. nakaSeptember 20, 2013 at 10:07 PM

    Let EG and FH cut AB at P,Q respectively
    ∠BFQ = 1/2∠BFA = 1/2(90 - A/2 - B/2)
    ∠BQF = a + ∠BFQ = 90 + A/2 - B/2

    Now let ∠CBD = x, ∠CDB = y, so that x+y=a.
    ∠PEA = 1/2∠BEA = 1/2∠(180 - y - (B-x)) = 90 - y/2 - B/2 + x/2
    ∠BPE = ∠PEA + y = 90 - B/2 + x/2 + y/2 = 90 - B/2 + A/2

    So ∠BPE=∠BQF
    q.e.d.

    ReplyDelete
  3. Emil EkkerOctober 2, 2013 at 2:59 AM

    See following picture:

    http://img40.imageshack.us/img40/9981/ptg.gif

    ReplyDelete
  4. Sumith PeirisFebruary 21, 2020 at 2:51 AM

    Let HF cut AC at X. Let < BFH = < AFH = q and let < FAC = < FBD = p

    Now < GEB = 1/2 < AEB = 1/2.(< BCE + < CBE) = 1/2.(p + 2q + p) = p + q = < FXC and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

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