Geometry Problem. Post your solution or view a solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Monday, January 11, 2016

### Geometry Problem 1176: Cyclic Quadrilateral, Diagonals, Six Diameters, Circles, Collinear Points, Concurrent Lines

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P16 and P26 lie on circle with diameter BD, We get Angle BP26P16 = Angle BDA

ReplyDeleteP12 and P26 lie on circle with diameter BC, We get Angle BP26P12 = Angle BCA

Since Angle BDA = Angle BCA , we have Angle BP26P16 = Angle BP26P12 , hence P16,P12 and P26 are colinear.

Similarly we can prove

P15,P14,P45

P25,P23,P35

P46,P34,P36 are colinear.

Lets assume radius of circle with diameter AC is R5, and radius of circle with diameter BD is R6.

We have AE.CE = BE.DE, we get R5^2 - O5E^2 = R6^2 - OE6^2, hence E must lie on common chord of circle 5 and 6 this proves P56,P56* and E are colinear.