Monday, June 8, 2009

Problem 298: Intersecting Circles, Chord, Secant, Midpoint, Congruence

Proposed Problem
Click the figure below to see the complete problem 298 about Intersecting Circles, Chord, Secant, Midpoint, Congruence.

 Problem 298: Intersecting Circles, Chord, Secant, Midpoint, Congruence.
See also:
Complete Problem 298
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

3 comments:

  1. AB and OQ are perpendicular bisector of each other. For two triangles OCD and QCE, obviously, a.s.s congruent to a.s.s..
    Angles D and E are acute angles because they are made by radius and secant. They are determined uniquely by law of sines. So they are congruent. Hence the two triangles OCD and QBE are congruent, and so are CD and CE.

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  2. Solving without using any trigonometry, connect AO, BO, AQ, BQ, CO, and QO. CO and QO are radii that bisects the same chord, so they are both perpendicular to AB, thus OCQ are collinear. AO=BO=AQ=BQ=r, Quadrilateral AQBO is a rhombus, CO=QO. Draw OF perpendicular to CD and QG perpendicular to CE; looking at Rt Triangles COF=CQG, because CO=QO, Angle OCF=Angle QCG, (AAS) so OF=QG, FC=GC. Connect OD and QE, OD=QE=r (hypotenuses), OF=QG (legs), so DF=EG.
    Since DF+FC=DC, EG+GC=EC, (substitution gets DF+FC=EC,) so DC=EC, therefore C is the midpoint of CE.

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  3. Sorry, I meant C is the midpoint of DE.

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