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# ICSE Class 9 Answered

Prove that a line joining the mid points of the parallel sides of a trapezium divides it into two equal parts
Asked by aswani1617 | 18 Sep, 2018, 11:02: PM
Expert Answer
G

Given : ABCD is a Trapezium
Point H and G are the mid point of parallel sides.

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

To prove : Area of AHGD = Area of HBCG

Let us assume that The given Trapezium is an isosceles Trapezium, with AD = BC

Here,
AH = HB and DG = GC      ....since Point H and G are the mid-point of AB and CD.

AHGE and  BHGF are rectagle.
(since AE = BF ..since AB || CD and  AE and BF are perpendicular to DG and GC respectively)

EG = AH .......(sides of a reactangle)
HB = GF ......(side of a rectangle)
Therefore Area of a AHGE =  Area of a BHGF ....(i)

In ΔADE and ΔBFC

AE = BF
AD = BC
and AED =  BFC = 900
Therefore ΔADE   ΔBFC   ...(RHS )
Therefore ΔADE = ΔBFC.......(ii)

Area of a AHGD = ΔADE +AHGE and
Area of a BHCG = ΔBFC + BHGF
from (i) and (ii), we get
Area of AHGD = Area of HBCG
Answered by Yasmeen Khan | 19 Sep, 2018, 02:37: PM

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