In the given figure, B and E are points on line segment AC and DF respectively
show that AD||CF.

Any one can Ansewer the Q |^|

0_{(opp angles of cyclic.quad) - (i)}

^{0(linear pair) - (ii)}

^{FROM (i) and (ii)}_'

SINCE⁰

^{SINCE ANGLES ON THE SAME SIDE OF A TRANSVERSAL ARE EQUAL, AD II CF}

CAD +BED =180(OPP ANGLES OF CYC QUAD)-i

BED + BEF= 180(LINEAR PAIR)-ii

FROM i AND ii,

CAD = BEF

BEF +FCA =180(OPP ANGLES OF CYC QUAD)

THEREFORE, FCA + CAD = 180 (SINCE CAD = BEF PROVED ABOVE)

SINCE THE CO-INTERIOR ANGLES (or) INTERIOR ANGLES ON THE

SAME SIDE OF THE TRANSVERSAL AC ARE SUPPLEMENTARY,

AD II CF