Question

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

Answer 1

$<b><u>ABCD is the trapezium H and G are the mid point of the parallel sides Construction: AE and BF are the height of the trapezium. To proof: Ar(AHDG) = Ar(HBCG) Assumption: Trapezium is an isosceles trapezium, with AD = BC AH = HB and DG = GC And EG = AH (sides of a rectangle) HB = GF (sides of a rectangle) So Ar(AHEG) = Ar(HBGF) In triangle ADE and BFC AE = BF AD = BC And angle  AED = angle BFC = 90 So ADE and BFC are congruent (RHS congruency) Hence Ar(ADE )and Ar(BFC) are equal Ar(AHDG) = Ar(ADE ) + Ar(AHEG) And Ar(HBCG) = Ar(HBGF) + Ar(BFC) Hence Ar(AHDG) =  Ar(HBCG) (proved)$