In Fig. 15.85, ABCD is a trapezium in which AB||DC and DC=40 cm and AB=60 cm. If X and Y are, respectively, the mid points of AD and BC, Prove that
(i) XY =50 cm (ii) DCYX is a trapezium
(iii) ar(trap. DCYX) = 
ar(trap. XYBA)
(i) Join DY and extend it to meet AB produced at P
∠BYP = ∠CYD (Vertically opposite angles)
∠DCY = ∠PBY (Since DC || AP)
BY = CY (Since Y is the mid-point of BC)
Hence, by A.S.A. congruence rule
ΔBYP ≅ ΔCYD
DY = YP
And,
DC = BP
Also,
X is the mid-point of AD
Therefore,
XY || AP
And,
XY = 
AP
XY = 
(AB + BP)
XY = 
(AB + DC)
XY = 
(60 + 40)
= 
× 100
= 50 cm
(ii) We have,
XY || AP
XY || AB and AB || DC
XY || DC
DCYX is a trapezium.
(iii) Since X and Y are the mid-points of AD and BC respectively
Therefore,
Trapezium DCYX and ABYX are of same height and assuming it as 'h' cm
Area (Trapezium DCYX) = 
(DC + XY) * h
= 
(40 + 50) h
= 45h cm2
Area (Trapezium ABYX) = 
(AB + XY) * h
= 
(60 + 50) * h
= 55h cm2
So,
= 
= 
Area of trapezium DCYX = 
Area of trapezium ABXY