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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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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
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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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
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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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
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi Bob
Thanks A Lot
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