Question

Prove that the lines joining the mid points of the parallel sides of trapezium divides it into two equal parts.

Answer 1

Answer:

Step-by-step explanation:

ABCD is the trapezium

H and G are the mid point of the parallel sides

Construction: AE and BF are the height of the trapezium.

To proof: Ar(AHDG) = Ar(HBCG)

Assumption: Trapezium is an isosceles trapezium, with AD = BC

AH = HB and DG = GC

And EG = AH (sides of a rectangle)

HB = GF (sides of a rectangle)

So Ar(AHEG) = Ar(HBGF)

In triangle ADE and BFC

AE = BF

AD = BC

And angle AED = angle BFC = 90

So ADE and BFC are congruent (RHS congruency)

Hence Ar(ADE )and Ar(BFC) are equal

Ar(AHDG) = Ar(ADE ) + Ar(AHEG)

And Ar(HBCG) = Ar(HBGF) + Ar(BFC)

Hence Ar(AHDG) = Ar(HBCG) (proved)