## Monday, August 26, 2013

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

Geometry Problem
Post your solution in the comment box below

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

Click the figure below to see the complete problem 915.

1. 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.

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3. angle(EFG)
= 360 - angle(DFE) - angle(DFG)
= (180- angle(DFE)) + (180 - angle(DFG))
= A + B
= 180 - C
= 180 - angle(GCE)
q.e.d.

4. ∠A + ∠DFE = 180°
∠B + ∠DFG = 180°

∠C + ∠EFG
= (180°−∠A−∠B) + (360°−∠DFE−∠DFG)
= 540° − (∠A + ∠DFE) − (∠B + ∠DFG)
= 180°

Thus, C,E,F,G concyclic.

5. Antonio,

I 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