Question

if the non parallel sides of a trapezium are equal prove that it is cyclic

Answer 1

construction: draw a line II to AC through D meeting AB extended at E.

AB II CD (given)

=> AE II CD, also AC II DE (const.)

So ACDE is a IIgm.( since both pair of opp. sides are parallel)

AC = DE (opp. sides of IIgm are equal)

AC = BD (given)

So BD = DE, From isos. tr. prop. we can infer that <DBE = <DEB .........1

also <ACD = <AED ( opp. andles of IIgm) ..................... 2

<DBE = <ACD ( from 1 and 2) .............. 3

also <DBE = <BDC (alternate int. angles) ................ 4

From 3 and 4, <ACD = <BDC ...................5

<ACD + <CAB = 1800 ( allied angles)

<BDC + <CAB = 1800 ( from 5)

Since opposite angles of the quadrilateral are supp. it is a cyclic quadrilateral.

Answer 2

in a cyclic trapezium base Angles are equal