If A, B, C, D are four points on a circle in order such that AB = CD, prove that AC = BD.

Question

If A, B, C, D are four points on a circle in order such that AB = CD. How do you prove that AC = BD.

Te problem is I do not even know how to start with his problem. If I can be given a hint on how to start, I will try to figure this out.. — Indu, Feb 29 '16 at 14:48
It may or may not be useful here, but in general, if the quad has 4 points on a circle, then the opposite angles always add up to 180° — imranfat, Feb 29 '16 at 14:55

score 2 · Accepted Answer · edited Feb 29 '16 at 15:18

2

$AB=CD$ and $AO=BO=CO=DO=R \implies$
$\triangle AOB \cong \triangle COD \implies$
$\angle AOB = \angle COD$

$\angle AOC = \angle BOD$ and $AO=OC=OB=OD\implies$
$\triangle AOC =\triangle BOD \implies$
$AC=BD$

edited Feb 29 '16 at 15:18

Ben Grossmann

answered Feb 29 '16 at 14:53

Roman83

score 2 · Answer 2 · answered Feb 29 '16 at 15:32

2

Let $M$ be the midpoint of the arc $BC$. Then the complete figure is symmetric with respect to the line $O\vee M$.

answered Feb 29 '16 at 15:32

Christian Blatter

2 Answers2