Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem.Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 1002.
Denote 3 angles of triangle ABC as α, β, γLet FB cut circle DBC at G’, We will prove that EG’ // to AB=> G coincide to G’As EF//BC we have ∠(ACB)= ∠(AEF)= γSince BDCG’ is cyclic => ∠(ACB)= ∠(BG’D)= ∠(FG’D)= γ => FDEG’ is cyclicSince FBDA is cyclic => ∠(BAD)= ∠(BFD)= αIn triangle DFG’ since ∠(DFG’)= α and ∠(FG’D)= γ so ∠(FDG’)= ∠(FEG’)= βAngle (CEG) supplement to ∠(DEG’)= β+ γ so ∠(CEG)= α => EG’//ABSo point G’ coincides to G and DEGF is cyclic
>We will prove that EG’ // to ABPeter, how to prove?
See correction of last lines below due to typo error:Angle (CEG') supplement to ∠(DEG’)= β+ γ so ∠(CEG')= α => EG’//ABSo point G’ coincides to G and DEGF is cyclic
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