9Given: KLMN is a quadrilateral.
 
KN=LM, angle KNM = angle LMN.
 
To prove: That the quadrilateral is cyclic, i.e
 
All points lie on the circle.
 
Proof: Side KN = Side LM.......................{Given}
 
Therefore, opposite sides of the quadrilateral are equal.
 
Therefore, the quadrilateral KLMN is a Rectangle.
 
Now,
 
A quadrilateral is cyclic only when their opposite angles form 180°
 
But all angles of a rectangle are 90°
 
Therefore,
 
Angle KNM + Angle KLM
 
90° + 90°
 
180°
 
Therefore, quadrilateral KLMN is cyclic.
 
Therefore points K, L, M, N lie on the circle.
 
 
						

							
dear student, note that aruns answer is totally rubbish, misleading and WRONG.
draw the fig KLMN. 
join KM and LN.
now, angle KNM= angle LMN (given)
NM= MN (common side)
KN= LM (given)
by SAS, tri KNM is congruent to tri LMN.
by CPCT, KM= LN......(1)
now, KL= LK (common side)
KN= LM (given)
LN= KM (from 1)
hence by SSS, tri KNL is congruent to tri LMK.
by CPCT, angle NKL= angle MLK= x (say).
also, given angle KNM= angle LMN= y (say)
so that x+x+y+y= 360
or x+y= 180
or angle MLK + angle KNM= 180.
since the sum of opposite angles of a quad being 180 deg implies its cyclic nature (converse of cyclic quad theorem), hence we deduce that the points k,l,m and n lie on a circle.
further, note that such a quad in general shall be a trapezium, not a rectangle as arun mentioned.
KINDLY APPROVE :))