Question

ABCD is a kite in which AB = AD and BC = DC. M, N and O are mid-points of sides AB, BC and CD. Prove that
(i) ZMNO = 90°

(ii) The line MP drawn parallel to NO bisects AD. ​

Answer 1

(i) If x = - 1/4 and y = - 7/4 then

(a) (x - y) > 0

(b) (x - y) < 0

(c) (x - y) = 0

(a) 2/5

(b) -10

(c) 4

(a) 1/12

(b) 1/3

x = 5/8 (iv) If and y = - 5/4 then

(a) (x + y) > 0

(b) (x + y) < 0

(c) 1/2

(d) 1/8

(d) (x + y) >= 0

(c) (x + y) = 0

(d) - 1/15