ABCD is a trapezium in which AB is parallel to DC. DC=30 cm. and AB=50cm. if X and Y are respectively the mid points of AD and BC, prove that (Nine times the area of DCYX= seven times the area of XYBA.)
Here is the answer to your query.
Given : ABCD is a trapezium with AB || DC
Construction : Join DY and produce it to meet AB produced at P.
In ∆BYP and ∆CYD
∠BYP = ∠CYD (vertically opposite angles)
∠DCY = ∠PBY (∵ Alternate opposite angles as DC || AP and BC is the transversal)
and BY = CY (Y is the mid point of BC)
Thus ∆BYP ≅ ∆CYD (by ASA cogence criterion)
⇒ DY = YP and DC = BP
⇒ Y is the mid point of AD
∴ XY || AP and XY = AP (mid point theorem)
⇒ XY = AP = (AB + BP) = (AB + DC) = (50 + 30) = × 80 cm = 40 cm
Since X and Y are the mid points of AD and BC respectively.
∴ trapezium DCYX and ABYX are of same height, say h cm
⇒ 9 ar(DCXY) = 7 ar(XYBA)