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

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

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## Sunday, July 12, 2015

### Geometry Problem 1133: Circle, Chord, Tangent, Circumcircle, Similarity, Equal Product

Labels:
chord,
circle,
circumcircle,
similarity,
tangent

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http://s17.postimg.org/5cijbp9e7/Pro_1133.png

ReplyDeleteDraw line BMN// GH ( M on EC and N on GC)

Since GH tangent to circle EFC so

∠ (EDF)= ∠ (GEF)= ∠ (BEH)=∠ (MBE)

Triangle FED similar to MEB ( case AA)

So EM/EF= EB/ED ……..(1)

Triangle AED similar to CEB ( case AA)

So EC/EA=EB/ED…… (2)

From (1) and (2) we have EM/EF=EC/EA => MF//CA

And FA/FB= MN/MB …..(3)

Since NB//GH => EG/EH=MN/MB…..(4)

From (3) and (4) we have FA/FB= EG/EH

<BEH=<GEA=<FDE, so <CHE=<CBE-<BEH=<ADE-<FDE=<AFD ( 1 ).

ReplyDeleteLet the circle (CGH) intersect CE at P and, easily, triangles GPH and ADB are similar and <GPE=<EHC(=<ADF as per (1)), consequently DF and PE are homologous lines within the 2 similar triangles, thus solving our problem.