Thursday, August 12, 2010

Problem 500: Circle, Diameter, Chord, Perpendicular, Measure

Geometry Problem
Click the figure below to see the complete problem 500 about Circle, Diameter, Chord, Perpendicular, Measure.

Problem 500. Circle, Diameter, Chord, Perpendicular, Measure.


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Complete Problem 500

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 5:48 PM
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2 comments:

  1. VihaanAugust 13, 2010 at 8:33 PM

    /_CDB=/_CAB from concyclic ABCD but /_CAB=/_BDF from concyclic AFED. Thus, /_CDB=/_BDF or BD bisects /_FDC internally. Likewise, CA bisects /_DCF internally. So FE bisects /_DFC or E is the incentre of Tr. Consider Tr. DFG where FE is the internal bisector. Now FB is perpendicular to FE so FB is the external bisector of /_ DFG. Hence, BG/BF=3/5 or x/(x+8)=3/5 giving us x=12.
    Vihaan, Dubai

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  2. chamacheSeptember 30, 2010 at 5:45 AM

    Se demuestra facilmente que el punto E es el ortocentro del triangulo formado por AE y las proplongaciones de los lados AD y BC luego el triangulo DFC el punto E es su incentro luego los puntos D, E , G y B son armonicos porlo tanto
    x= 12
    wirocha

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