Geometry Problem

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## Friday, January 14, 2011

### Problem 566: Quadrilateral, Diagonals, Parallel

Labels:
diagonal,
parallel,
quadrilateral

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▲EOB ~ ▲OCD => EO/OC = OB/OD

ReplyDelete▲AOB ~ ▲OCF => AO/OC = OB/OF

=>

AO/EO = OF/OD

from similarity or Thales theorem

EF//AD

Let I is the intersection of AC and BD

ReplyDelete1. Triangle IAB similar to IFC ( Case AA)

And IF/IB=IC/IA

So IF/ID=IF/IB * IB/ID= IC/IA * IB/ID (1)

2. Triangle IBE similar to IDC ( Case AA)

And IE/IC=IB/ID

So IE/IA=IE/IC *IC/IA= IB/ID *IC/IA (2)

3. compare (1) to (2) we have IF/ID= IE/IA and triangle IEF will similar to IAD (case SAS)

And EF// AD

Peter Tran