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

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

Click the figure below to enlarge the problem 947.

## Thursday, December 19, 2013

### Geometry Problem 947: Triangle, Circumcircle, Angle Bisector, Chord, Arc, Parallel

Labels:
angle bisector,
arc,
chord,
circumcircle,
parallel,
triangle

Subscribe to:
Post Comments (Atom)

Angle ABE = angle AFE = angle GBD

ReplyDeleteSo BDFG concylic

Angle FAC = angle FBG = angle FDG

QED

Let AF and BC intersect at P.

ReplyDeleteSince ABFC concyclic, (AD+DP)×FP = (CG+GP)×BP ... (1)

Since BDGF concyclic, DP×FP = GP×BP ... (2)

(1) - (2): AD×FP = CG×BP ... (3)

(2) / (3): DP/AD = GP/CG

Hence, DG//AC.

BDGFconcyclic=>< BFD= DG||AC (formeazacu secanta BC unghiuri corespondente congruente)

ReplyDeleteBFD=BGD,BFA=BCA, BGD=BCA,=> DG||AC

Delete