Geometry Problem. Post your solution in the comments box below.

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

Click the figure below to enlarge the problem 951.

## Friday, December 27, 2013

### Geometry Problem 951: Intersecting Circles, Chord, Diameter, Parallel, Cyclic Quadrilateral, Concyclic Points

Labels:
chord,
concyclic,
cyclic quadrilateral,
intersecting circles,
parallel,
tangent

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2(180°−∠EAD)=2∠AFE=∠AOE=∠OAC=90°−∠CAD

ReplyDelete=∠QAD=∠AQG=2∠AHG=2∠DAH

Thus, E,A,H are collinear.

Similarly, F,A,G are collinear.

By the above argument,

∠EFG=∠EFA=∠AHG=∠EHG

Hence, F,E,H,G concyclic.

<CEA=<DGA=<CAD. <DAH=45-<CAD/2, and <EAD=90+<CAD+(45-<CAD/2), so <EAD+<DAH=180 and EAH are collinear. Same reasoning for FAG to be collinear. Then <EFG=<EHG=45-<CAD/2.

ReplyDeleteFrom the give ͡ ABC = ͡ ABD => ˂FEH = ˂FGH

ReplyDelete