Start with any parallelogram ABCD, and construct squares on its sides, as shown in the diagram below. Points P, Q, R, and S are the intersections of the diagonals of the squares. By using only the steps provided, prove that PQRS is also a square.

a) Why is triangle ABH congruent to triangle DCK?

b) Show that AP = DR

c) Why is AQ = QD?

d) Prove angle PAQ is congruent to RDQ (hint: here we need to angle chase a little bit. start by trying this: let measure of angle ADC = x, and figure out angle GAF also has measure x)

e) Why is triangle PAQ congruent to RDQ?

f) Now we have segment PQ congruent to segment QR and angle DQR is congruent to angle AQP...why? (should include CPCFC)

g) Now prove measure of angle PQR = 90

h) So, we have seen that the triangle congruence in part (e) ultimately leads to segment PQ is congruent to segment QR and measure of angle PQR = 90. Similarly, there is a triangle congruence that would give you line segment QR congruent to line segment RS and measure of angle QRS = 90. Which one? Also, which triangle congruence gives you line segment RS congruent to SP and measure of angle RSP = 90? (note: don't prove anything there, since the proofs are exactly similar to what we did earlier in this problem. Just tell which congruences you would need to reach these conclusions)

i) Finish the proof.

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a) Why is triangle ABH congruent to triangle DCK?

We started with a parallelogram so AB=DC.

We put squares on the side, so AB=BH and DC=DK.

Since ABH and DCK make right angles with the parallelogram the triangles ABH and DCK are congruent.

b) Show that AP = DR

We show that the triangles ABP and DCR are congruent.

P is the intersection of the diagonals of the square on side AB.

Therefore angle PAB=angle PBA = 45 degrees

R is the intersection of the diagonals of the square on side DC

Therefore angle RDC=angle RCD = 45 degrees

Since AB=DC, the triangles ABP and DCR are then congruent.

This means that AP=DR....