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In a convex quadrilateral PQRS, triangles PQR and PSR have the same area. E is the intersection of the diagonals of the quadrilateral. How do I prove that the parallels through E to lines PS, SR, RQ and QP meet PQ, QR, RS and SP at midpoints? In other words how can I say that the the parallel through PS meets PQ at its midpoint? I know i need to use the fact that the area of the triangles PQR and PSR are the same, can you expand?

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    I edited my answer... – 2017-02-18

2 Answers 2

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Consider the following figure.

enter image description here

The key to the solution is the observation that the two red triangles are congruent. That is the crossing point (red) halves $PR$. So does it with any segment connecting the two white parallel lines. (It halves even the blue segment having nothing to do with our quadrilateral.)

EDIT

Take a look at the following figure:

enter image description here

Assume that we have the thick red segment and its mind point. Now draw a line through the upper point of the red segment. Then draw parallels to this line through the mid pint and through the lower point. (white lines) Then draw the thick blue segment joining the outer white parallels. I claim that the inner parallel line halves the thick blue segment.

To prove this statement construct a parallel to the thick red line through the crossing point of the whit middle parallel and the thick blue line. (broken red line) Then draw a parallel to the thick blue line through the mid point of the thick red segment. (broken blue) So many angles and so many segments equal that we can tell that the two "hour glasses" are congruent. That is, the middle white line halves not only the thick red segment but the thick blue segment as well.

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    The two triangle (upper and lower) has a common base and a common area. As a result their height are the same. The red triangles have the same height, these heights are perpendicular to the common base and they have one more common angles at the midpoint. – 2017-02-18
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    One common edge, two common angles next to the common edge opposite to the common edge. – 2017-02-18
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Let $X$ and $Y$ be the feet of the perpendiculars from $S$ and $R$ onto the line $PR$ respectively. Since the areas are of the given triangles are equal, we have $SX=RY$ and it follows that trisngles $SXE$ and $RYE$ are congruent.

Therefore $E$ is the midpoint of $SQ$ and all the stated results will then follow by similar triangles.

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    $QY=SX$, angles $QYE=SXE=90$ and angles $QEY=SEX$ are opposite angles so congruent triangles – 2017-02-18
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    In each case use similar triangles – 2017-02-18