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Quadrilateral $PQRS$ is a trapezium ($PQ||SR$ and $PQ>SR$).

Points $X$ and $Y$ are mid points of diagonal $PR$ and $QS$ respectively.

Prove that segment $XY$ parallel to side $PQ$ and $XY= \frac{1}{2}(PQ-SR)$.

I tried this question whole day but couldn't find answer please help me in doing it

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    http://www.meritnation.com/ask-answer/question/prove-that-the-line-segment-joining-the-mid-points-of-the-di/quadrilaterals/1733376 – 2017-02-11
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    that is a very known result called "Euler average". It is a imediate consequence of other two very known result called "average base" in a trapezium and $average base$ in a triangle. – 2017-02-11

2 Answers 2

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Draw the line passing through point $Q$ and parallel to edge $PS$ and let that line intersect the line $SR$ at point $N$ (see picture). Then $PQNS$ is a parallelogram because edge $SN$ is parallel to edge $PQ$ and edge $PS$ is parallel to edge $QN$. Consequently, the intersection point of the diagonals $PN$ and $QS$ is the midpoint of $QS$ (and $PN$), which is point $Y$. Furthermore, $XY$ is a mid-segment in triangle $NPR$ parallel to $RN$ and half of its size. This means that $XY$ is parallel to $PQ$, because $PQ$ is parallel to $RN$, and $XY = \frac{1}{2} RN$. However, $RN = SN - SR$ and since $SN = PQ$ we get $$XY = \frac{1}{2} (PQ - SR) \,\,\, \text{ as well as } \,\,\, XY \, || \, PQ$$

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Let $a||PQ$ and $X\in a$.

Let $a\cap QR=\{M\}$ and $a\cap SP=\{N\}$.

Since $X$ a middle point of $PR$, we get that $M$ is a middle point of $QR$

and since $PQRS$ is trapezium, we get that $N$ is a middle point of $PS$ and $Y\in a$.

Thus, $XY||PQ$ and $$XY=NY-NX=\frac{PQ}{2}-\frac{SR}{2}=\frac{PQ-SR}{2}$$