Prove that the segment joining the middle points of two non-parallel sides of a trapezium is parallel to the parallel sides and half of their sum.

Explanation

Given that ,
In the trapezium `ABCD`,
`AD |\| BC, AX = XB` and `DY = YC`
We have To Prove:
`(i) XY |\| AD` or `XY |\| BC`
(ii) `XY = 1/2(AD + BC)`
we have to Extend BA and CD to meet at Z.
and Join A and C.
Let it cut `XY` at `P`
`(i)`according to question ,
In `△ZBC,AD∣∣BC `
`therefore (ZA)/(AB)=(ZD)/(ZC)`
`=>(ZA)/(2AX)=(ZD)/(2DY)`
`=>(ZA)/(AX)=(ZD)/(DY)`
`=>XY |\| AD`
`(ii)` In `△ABC,AX=XB`
`XP∣∣BC `
`∴AP=PC`(∵ Converse of mid point theorem)
`∴XP=1/2BC`
In `△ADC,PY=1/2AD`
By adding, we get ` XP+PY=1/2BC+1/2AD`
`=>XY=1/2(BC+AD)`