Geometry Problem

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

Click the figure below to see the complete problem 915.

## Monday, August 26, 2013

### Problem 915: Triangle, Intersecting Circles, Concyclic Points, Cyclic Quadrilateral

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Join D to F and extend it to D1. ///ly join G to F and extend to G1. Since ADFE is concyclic, /_EFD1 =/_A. ///ly /_GFD1=/B which implies that /_EFG1=/_C since A,B & C are internal angles of Tr. ABC. This, in turn, means that quad CEFG is concyclic.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteangle(EFG)

ReplyDelete= 360 - angle(DFE) - angle(DFG)

= (180- angle(DFE)) + (180 - angle(DFG))

= A + B

= 180 - C

= 180 - angle(GCE)

q.e.d.

∠A + ∠DFE = 180°

ReplyDelete∠B + ∠DFG = 180°

∠C + ∠EFG

= (180°−∠A−∠B) + (360°−∠DFE−∠DFG)

= 540° − (∠A + ∠DFE) − (∠B + ∠DFG)

= 180°

Thus, C,E,F,G concyclic.

Antonio,

ReplyDeleteI would like to propose a new problem:

D is located inside the triangle ABC, such that <DAC=<DAB=<DBA=10 degrees.

<DBC is 30 degrees. Find <ACD.

Respectfully,

Erina-NJ