The applet suggests the following construction of a square:

The proof below assumes that P and Q lie between A and R. This need not necessarily be the case, but if P and Q are without AR, the proof has to be slightly adjusted.

First consider right triangles APD and AQB:

It follows that ΔAPD = ΔAQB. In particular, AD = AB and ∠ADP = ∠BAQ. Therefore, ∠BAD is right.

Further, define T on the extension of RC such that BT||AR and consider ΔBTC.

and

Hence ΔBTC equals the other two. In particular, BC = AB and ∠ABC is also right. This does show that ABCD is a square.

References