# Conic from Parallel Chords

### What Might This Be About?

### Problem

Two circles intersect at points $A$ and $B.$ Chords $EF$ on one and $GH$ on the other are parallel.

Prove six points $A,B,E,F,G,H$ lie on a conic.

### Solution

The easiest way to prove the statement is to show that it is equivalent to its inverse. This is done in a reasonably standard manner by *reductio ad absurdum*.

### Converse

Given six points on a conic, $A,B,E,F,G,H.$ Such that $ABFE$ and $EBHG$ are cyclic quadrilaterals. Then $EF\parallel GH.$

### Acknowledgment

The problem has been posted by Dao Thanh Oai at the CutTheKnotMath facebook page.

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