## Thursday, August 5, 2010

### Problem 492: Cyclic Quadrilateral, Circle, Perpendicular, Parallel

Geometry Problem
Click the figure below to see the complete problem 492 Cyclic Quadrilateral, Circle, Perpendicular, Side, Diagonal, Parallel.

Complete Problem 492

Level: High School, SAT Prep, College geometry

1. Denote ( XYZ)= angle XYZ
1. Connect EC and BG
Note that E,O,C are collinear ( EAC)=90
B,O,G are collinear (BDG)=90
Quadrilateral EGCB have diagonals intersect at midpoints so EGCB is a parallelogram and BC//EG
And (DFH)=(DAH) ( both angles face the same arc DH)
But (DAH)=(DBC)…………...( both angles face the same arc DC)
So (DFH)=(DBC) (corresponding angles)
And BC// FH

Peter Tran
vstran@yahoo.com

2. 1) GDC = 90 - BDC, EAB = 90 - BAC => GDC = EAB =>
arc GC = arc EB => GE // BC
DBC = DAC, DFH = DAC => DBC = DFH as corresponding ang
=> BC // FH

3. Since ADHF is cyclic < HFD = < HAD = < CBD hence BC//FH

Also external < H = < EAD = < EGH hence EG //FH

So FH // BC // EG

Sumith Peiris
Moratuwa
Sri Lanka