Tuesday, June 29, 2010

Problem 468: Circle, Chord, Midpoint, Congruence, Angle

Geometry Problem
Click the figure below to see the complete problem 468 about Circle, Chord, Midpoint, Congruence, Angle.

Circle, Chord, Midpoint, Congruence, Angle
See also:
Complete Problem 468

Level: High School, SAT Prep, College geometry


  1. OP^2=OE^2-PE^2=OG^2-QG^2=OQ^2 since PE=QG (because of equal chords) & OE=OG (both=radius of the circle). Thus,OP=OQ.Similarly,OP=OQ=OM=ON or M,P,Q & N are concyclic; hence α = β

  2. Note that OM perp. to AB; OP perp. to EF ; OQ perp. to GH; ON perp. to CD (property of midpoint of a chord)
    and OM^2= OP^2=OQ^2=ON^2=Radius^2 –(.5 Chord length) ^2
    so OM=OP=OQ=ON and MPQN is cyclical quadrilateral
    and angle (alpha)= angle (beta) ( 2 angles MPN and MQN face the same chord )

    Peter Tran

  3. Equal chords of a circle are equidistant from the centre. Hence PQNM lie on a circle whose centre is O and the result follows

    Sumith Peiris
    Sri Lanka