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.

See also:
Complete Problem 500

Level: High School, SAT Prep, College geometry


  1. /_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

  2. 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