Friday, August 6, 2010

Problem 493: Right Triangle, Circumcircle, Perpendicular, Chord

Geometry Problem
Click the figure below to see the complete problem 493 about Right Triangle, Circumcircle, Perpendicular, Chord.

Problem 493. Right Triangle, Circumcircle, Perpendicular, Chord.
See also:
Complete Problem 493

Level: High School, SAT Prep, College geometry


  1. Denote (XYZ) =angle XYZ
    Connect AD and note that quadrilateral AFED is cyclic since (AFD)=(AED)=90
    We have (EFD) = (EAD) ( angles face the same arc ED)
    (EAD)=(CBD) (angles face the same arc CD)
    (CBD)=(BDF)…………(corresponding angles)
    So (EFD)=(BDF)
    And ( DBF)=(BFG) (both angles complement to (EFD)=(BDF)
    Triangles FGD and BGF are isosceles and GD=GF=GB
    In isosceles triangle BOD , median OG will perpendicular to the base BD

    Peter Tran

  2. We can use the simpson line !

  3. Adil

    Yes, Simpson line is the better solution !
    Thanks Adil.

  4. < EFD = < EAD = < DBC = < BDF

    Hence in right triangle BFD, G is the centre and the result follows

    Sumith Peiris
    Sri Lanka

  5. Join AD and let m(DAC)=X
    => m(ABD) = 90-X -----(1)
    Since BFD is right triangle, m(BDF)=X ------(2)
    Observe that AFED are concyclic => m(EFD) = m(EAD) = X ----------(3)
    From (2) and (3) we can conculde that the triangle GFE is isosceles and GF=GD ---------(4)
    Since m(BFD)=90 and from (4), it can be said that G is the circum-center of the right triangle BFD and hence BG=GD
    Hence OG is perpendicular to the chord BD at its center G