Geometry Problem

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

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Complete Problem 468

Level: High School, SAT Prep, College geometry

## Tuesday, June 29, 2010

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

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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

ReplyDeleteNote that OM perp. to AB; OP perp. to EF ; OQ perp. to GH; ON perp. to CD (property of midpoint of a chord)

ReplyDeleteand 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

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

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka