PQ is diameter . Show that QC & AP bisect angle C & angle A.

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- February 26th 2013, 07:01 AMsreesujaGeometry
PQ is diameter . Show that QC & AP bisect angle C & angle A.

- February 26th 2013, 08:16 AMibduttRe: Geometry
Pl Give some information about A and C

- February 26th 2013, 09:02 PMsreesujaRe: Geometry
ABCD is a cyclic quadrilateral.where PQ is the diameter. Show that QC and AP bisect angle C & and angle A.

- February 26th 2013, 10:23 PMibduttRe: Geometry
I find stii something missing. Please see attached figureAttachment 27268

- February 26th 2013, 11:23 PMsreesujaRe: Geometry
The diagram is

- February 27th 2013, 12:19 AMMINOANMANRe: Geometry
Sreesuja

the problem you mentioned is wrongly formulated.

it must be like this .

In a cyclic quadrilateral ABCD the angle bisector of A meets the circle at P and the angle bisector of C meets he circle at Q .Show that PQ is a diameter of the circle.

Be carefull the converse you are doing is not correct ..not all the diameters of the circle are bisectors of the angle C or A!!!!. - February 27th 2013, 12:53 AMMINOANMANRe: Geometry
Proof:

Please revise your diagram also it is wrong .put in the place of A Q AND IN PLACE OF Q... A ALSO INTERCHANGE C WITH P

THEN USE THE MAIN PROPERTY OF CYCLIC QUADRELATERALS..THAT STIPULATES : angle A + angle C =180.

here half of angle A + half of angle C must be 90.

Indeed look: the upper half of A has a DP as the coresponding arc. similarly the upper half of C has as coresponding arcs QA+AD

THEREFORE ARC(QA)+ARC(AD)+ARC(DP) = 180 AND THE POINTS Q AND P LYING ON A DIAMETER.

AS SIMPLE AS SUCH

MINOAS