Answers

2015-11-30T20:55:37+05:30
Let ABCD be the cyclic trapezium with AB IICD.
thru' C draw CE parallel to  AD meeting AB in E
.SoAECD is a parallelogram.
so
angle D=angle AEC.... opp angles of a parallelogram are equal....(i)
but
angle D+angle ABC=180... opp angles of a cyclic quadr are supplementary....(ii
from (i) and (ii)
angle AEC+angle ABC=180
but triangle AEC+angle CEB= 180...linear pairso angle
ABC= angle CEB ..(iii)
so CE=CB...  sides opp equal angles are equal.(iv) 
butCE=AD...opp  sides of parallelogram AECD.from (iv)
we get
AD=CB
Thus cyclic quadri ABCD is isoceles.
this proves the first part of the question.
now, join AC and BD, the diagonals
.in triangles DAB and CBA,AD=CB...proved before
AB=AB commonangle ADB= angle ACB..
 angles in the same  segment of a circle are equal
.here AB is the chord.so triangles DAB and CBA are congruent....SAS rule.
so AD=CB... CPCT 
hence proved.
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