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**shivusuja****Member**- Registered: 2006-09-14
- Posts: 56

Two circles intersect at P & Q. Through P , two straight lines APB & CPD are drawn to meet the circles at A , B, C,& D. AC & DB when produced meet at O. Show that OAQB is cyclic quadrilateral

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,531

Hi

Have no access to my geometry software at the moment.

Maybe joining PQ and playing around with equal angles made by same chord will work.

I will come back to this when I am able.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,531

Like this

A C D = A Q P

O D C = P Q B

So A O B add A Q B = 180

Hope you can fill the gaps

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,531

Additional:

Since Yesterday Iam Strugling With That. Does The Point O Comes Out Of The Circles. How Is ACD= AQB.

Iam Find ACD+AQB=180. Please Can You Give More Clarity

Yes, point O is outside the circles.

ACD is not = to AQB.

I said ACD = AQP

With AP as a chord, angle ACP = angle AQP because they are two angles on the circumference made by the same chord.

ACP = ACD because C-P-D is a straight line.

Hence ACD = AQP

Then

QPDB is cyclic so PQB + PDB = 180.

But on the straight line ODB, angle ODC + PDB = 180

therefore, angle ODC = angle PQB.

So we have in triangle OCD,

OCD + ODC + DOC = 180 => ACD + PQB + DOC = 180 => (AQP +PQB) + DOC = 180 => AQB + DOA = 180

So OAQB is cyclic.

Hope that clears it up.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**shivusuja****Member**- Registered: 2006-09-14
- Posts: 56

Hi Bob

Thanks A Lot

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