# Geometry proof

1. Dec 12, 2004

### Physics is Phun

I finished this proof that took about a whole page and I was just wondering if there were an easier way of doing things that would take less work. here's the problem.
PQ and PR are two chords of a circle with centre O. OT is perpendicular to PQ and OS is perpendicular to PR. IF OT = OS prove that T,S,R and Q are concyclic.
I am going to go through my steps without proof cause it would take forever. I just want to get an idea if I am doing it the easiest way.

It would probably help if you drew this out.

First I proved QT = RS
then QTO is congruent to RSO
Then you can prove TS is parallel to QR
from there i can conclude that <QSR = <QTR
from there can I conlude that they are concyclic by equal angles in a segment?

Edit: I realized that I didn't need to prove parallel lines. It really doesn't help the question in any way.

2. Dec 12, 2004

### Physics is Phun

Ya, as I look at it, I dont think this proof works. I think I assumed O was on QR which it doesn't have to be. Now I really need help !!

3. Dec 13, 2004

### Rogerio

T and S are not defined.
If you don't tell a little bit more about them, is gonna be impossible to prove anything.

4. Dec 15, 2004

### Popey

I suppose that T lies on PQ and S lies on PR
PS=SR
PT=TQ
This two equalities make the situation simple

At first we may notice that
If AB is a chord of a circle with centre O then the triangle OAB is an isoskeles, so the perpendicular from O to AB is at once
-perpendicular bisector of AB
-angle bisector of AOB
-altitude

The right triangles OPS,OPT are congruent (OS=OT, OP common)
So we get PS=PT
Then, it's easy to prove that the four triangles OPS,ORS,OPT,OQT are congruent
(actually there is a symmetry trough OP)

The way I use is, to prove that all perpendicular bisectors of RS,RQ,QT intersect each other at the point K.
Then KR=KS,KR=KQ,KQ=KT (so, the point K is the center of the circle we are looking for)

You can easy show that the perpendicular bisector on RQ is the line OP (using congruent triangles)

Take the midpoint A of RS, and respectively the midpoint B of TQ

bring the perpendicular bisector of RS, it intersects OP at the point K (theydo intersect since AK is parallel to SO and SO intersects PR)

You have the right triangle AKP,

from K bring the perpendicular line to PQ, which intersects PQ at the point B
Now, you have another right triangle BKP
But their angles BPK and APK are equal => Triangles AKP,BKP are similar
If you use that KP is common, you get that AKP and BKP triangles are congruent

So KA=KB
and you can use it to prove KR=KS=KT=KQ

I hope that helps

You can simplify further the proof if you use symmetry
Everything over OP is symmetric to everything on the other side of OP