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## Answers

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.