## Sunday, April 6, 2014

### Math: Geometry Problem 1002: Triangle, Circumcircle, Cevian, Parallel Lines, Cyclic Quadrilateral, Concyclic Points

Geometry Problem.

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

Click the figure below to see the complete problem 1002. 1. Denote 3 angles of triangle ABC as α, β, γ
Let FB cut circle DBC at G’, We will prove that EG’ // to AB=> G coincide to G’
As EF//BC we have ∠(ACB)= ∠(AEF)= γ
Since BDCG’ is cyclic => ∠(ACB)= ∠(BG’D)= ∠(FG’D)= γ => FDEG’ is cyclic
Since FBDA is cyclic => ∠(BAD)= ∠(BFD)= α
In triangle DFG’ since ∠(DFG’)= α and ∠(FG’D)= γ so ∠(FDG’)= ∠(FEG’)= β
Angle (CEG) supplement to ∠(DEG’)= β+ γ so ∠(CEG)= α => EG’//AB
So point G’ coincides to G and DEGF is cyclic

1. >We will prove that EG’ // to AB
Peter, how to prove?

2. See correction of last lines below due to typo error:
Angle (CEG') supplement to ∠(DEG’)= β+ γ so ∠(CEG')= α => EG’//AB
So point G’ coincides to G and DEGF is cyclic

2. Angle (AFB) = Angle (BDC). So then Angle(BGC) = 180 - Angle(BDC) = 180 - Angle(AFB). Therefore, AF//GC. As a consequence, Angle(FAD)=180 - Angle(GCD). Then Angle(FBD)=Angle(GCD). As B, G, C and D are concyclical, Angle(DBG)=180 - Angle(GCD). Therefore, F, B and G are collinear.

As AB//EG, Angle(BAD)=Angle(GEC). As F, A, B and D are concyclical, Angle(BAD)=Angle(DFB).

Then, Angle(DFB)=Angle(GEC).What indicates that DEGF is cyclical cuadrilateral. W^5