Proving Perpendicular Lines in a Quadrilateral

[horsecandy911]

Homework Statement

ABCD is a quadrilateral with AB=CD
P,Q,R,S are the midpoints of AD,AC,BD,BC respectively
Prove that PS is perpendicular to QR

Homework Equations

The Attempt at a Solution

I've tried a bunch of different avenues, but I think that the most promising one is to somehow show that triangle PQR or SQR is isosceles; then, if we can also prove that PS is the angle bisector or the median of PQR, we will also know its an altitude, so it must make a right angle with QR. But I don't know how to show that PQR or SQR is isosceles!

 

[epenguin]

Not extremely obvious - the figure is quite cluttered so hard to see. I couldn't see it through to the answer yesterday but this morning was clearer. Yesterday I thought similar triangles would be useful and they are. Your problem needs you to call on one of the handful of Polya 'How to solve it' principles I have mentioned in other threads, this one is: 'Are you using all the information you are given?'.

You can say something about those lengths you mention and you are along exactly the right lines.

 

[horsecandy911]

Aha! Thank you epenguin, you nudged it over the edge for me. For anyone who's been trying the problem and wondering, here is the solution:

AQ is 1/2 AC because Q is the midpoint of AC
AP is 1/2 AD because P is the midpoint of AD
and triangles APQ and ACD share angle PAQ
So triangles APQ and ACD are similar
Therefore PQ = 1/2 CD
You can make an exactly analogous argument for the other side to show that
PR =1/2 AB
And since AB = DC
PR = PQ
Make the same argument for the opposite side to obtain that SR and SQ = 1/2 CD and AB respectively
So because AB = DC again, SR = SQ = PR = PQ
So PQRS is a parallelogram and its diagonals bisect each other,
So PS crosses QR at the midpoint of QR,
Making it the median of triangle PQR;
And the median of an isosceles angle from its vertex is the same as its altitude, so PS is an altitude of triangle PQR, which meets QR at 90 degrees,
Therefore PS and QR are perpendicular

Whew... that seemed rather long and complicated... anyone know if I'm missing some more elegant solution?

 

[epenguin]

Not much but once you are past the similar triangles I found it necessary to talk only of lengths, not angles. You might now look up or re-read 'rhombus' in your textbook and you may get some reinforcement or even now feel you are coming to it from on top whereas previously you came from underneath.