See complete Problem 153

Circumscribed Quadrilateral, Diagonals Concurrent with Chords. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Sunday, August 3, 2008

### Elearn Geometry Problem 153

Subscribe to:
Post Comments (Atom)

AB cut CD at O and OE=OG, DH/HA*AE/EO*OG/GD=1, so BD,FH,CA concur, similarly, so is EG.

ReplyDeleteThe above proof is wrong!!

ReplyDeleteEG,FH cuts BD at M,M'

referring to Proposed Problem 152

BM'/M'D=BF/HD=BE/GD=BM/MD, M=M'

A more "explicit" solution to problem 153.

ReplyDeleteLet’s admit that EG meets BD at M, and FH meets BD at M’. By problem 152, BE/GD = BM/MD, or BF/HD = BM/MD. Also BF/HD = BM’/M’D, so BM’/M’D = BM/MD and M = M’. Thus EG, BD and FH are concurrent at M.

Similarly we can prove that EG, AC and FH are concurrent at a point which can only be M, because it’s the intersection between EG and FH.

Hence diagonals AC and BD are concurrent with chords EG and FH at M.

In AEBFCD, lines AC, BD and EG concur by Brianchon's theorem.

ReplyDeleteIn ABFCDH, lines AC, BD and FH concur by Brianchon's theorem.

So BD, AC, EG and FH concur.