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