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Topic review (newest first)

2013-03-01 15:50:40

Hi Bob
  Thanks A Lot

bob bundy
2013-03-01 06:54:56


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


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 bundy
2013-02-27 12:07:39

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 smile


bob bundy
2013-02-27 11:50:58


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.  sad


2013-02-27 02:14:23

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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