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Answer: hii mate so nice to help you i) Let ABCD is a trapezium, in which AB parallel to CD. E is the midpoint of BC.
ii) Through E draw a line parallel to parallel sides to intersect AD at F. Join AC, which intersects EF at G.
iii) In triangle ABC, EG parallel to AB and through the midpoint of BC;
So, by midpoint theorem of triangles, EG = (1/2)AB
Similarly GF = (1/2)CD
==> EF = EG + GF = (1/2)(AB + CD) ------------ (1)
iv) Draw an altitude from D on to AB, which meets EF at P and AB at Q.
==> Height of the trapezium ABCD = DP + PQ
Also heights of the triangles, FED and FEA are respectively DP & PQ.
v) Hence of these, area of ΔFED = (1/2)*EF*DP --------- (2)
and area of ΔFEA = (1/2)*EF*PQ ----------- (3)
Adding (2) & (3): Area of ΔFEA + Area of ΔFED = (1/2)*EF*(DP + PQ)
Substituting for EF from (1), sum of these areas = (1/2)*{(1/2)(AB + CD)*(DP + PQ)}
But (1/2)(AB + CD)(DE + PQ) = Area of trapezium ABCE
As well ΔFEA + ΔFED = Area of ΔAED,
Thus area of ΔAED = (1/2) of Area of trapezium ABCD ----------- (4)
vi By geometry of the figure,
Area ΔAEB + Area of ΔDEC = Area of trap. ABCD - Area of ΔAED
= Area of trap. ABCD - (1/2) of Area of trapezium ABCD = (1/2) of Area of trapezium ABCD
Thus Area ΔAEB + Area of ΔDEC = (1/2) of Area of trapezium ABCD [Proved]