Monday, August 24, 2009

Problem 347. Triangle, Altitude, Perpendicular, Circle, Concyclic points.

Proposed Problem
Click the figure below to see the complete problem 347 about Triangle, Altitude, Perpendicular, Circle, Concyclic points.

 Problem 347. Triangle, Altitude, Perpendicular, Circle, Concyclic points..
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Complete Problem 347
Level: High School, SAT Prep, College geometry

4 comments:

  1. BF BC = BD squared = BE BA,
    BF BC = BE BA
    By the converse of power theorem, AEFC is cocycle.

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  2. Quad. BEDF is clearly concyclic; hence angle BEF = angle BDF = 90 - angle FDC = angle C. Hence AEFC is concyclic. QED.
    Vihaan

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  3. angle BED + angle BFD = 180 degrees because DE and DF are perpendiculars to AB and BC respectively. this implies that BEDF is a cyclic quadrilateral. join EF then angle DEF = angle DBF as they are in the same segment. but angle DBF = 90 - angle C as BDC is a right triangle. so angle DEF = 90 - angle C. this implies that angle AEF = 90 + 90 - angle C = 180 - angle C. so angle AEF + angle ACF = 180. so AEFC is a cyclic quadrilateral because the sum of a pair of opposite angles equals 180 degrees. therefore points A, E, F and C are concyclic.
    Q. E. D.

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  4. Since EBFD is cyclic < BEF = < BDF which is in turn = to FCD.
    Hence AEFC is con cyclic

    Sumith Peiris
    Moratuwa
    Sri Lanka

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