Geometry Problem

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

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Complete Problem 492

Level: High School, SAT Prep, College geometry

## Thursday, August 5, 2010

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

Labels:
circle,
cyclic quadrilateral,
diagonal,
parallel,
perpendicular,
side

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Denote ( XYZ)= angle XYZ

ReplyDelete1. 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

2. Note that quadrilateral ADHF is cyclic (FAH)=(FDH)=90

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

1) GDC = 90 - BDC, EAB = 90 - BAC => GDC = EAB =>

ReplyDeletearc GC = arc EB => GE // BC

2) ADHF cyclic =>

DBC = DAC, DFH = DAC => DBC = DFH as corresponding ang

=> BC // FH

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

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

So FH // BC // EG

Sumith Peiris

Moratuwa

Sri Lanka