Assume that the length of a side of square ABCD is 4 units. Then

#BP=3#

#PC=1#

#QC=2#

#DQ=2#

#AD=4#

#AB=4#

Consider the right triangle #PQC# where angle #C# is 90 degrees. #PC# and #QC# are the legs of the right triangle. Since #PC=1# and #QC=2#, by the Pythagorean Theorem, the length of the hypotenuse #PQ=sqrt(1^2+2^2)=sqrt(5)#.

Now consider the right triangle #ADQ# where angle #D# is 90 degrees. #AD# and #DQ# are the legs of this triangle. Since #AD=4# and #DQ=2#, by the Pythagorean Theorem, the length of the hypotenuse #AQ=sqrt(4^2+2^2)=sqrt(20)#.

So

#(AQ)/(PQ)=sqrt(20)/sqrt(5)=sqrt(4)=2#.

Therefore

#AQ=2PQ#.

Now consider right triangle #ABP# where angle #B# is 90 degrees. #AB# and #BP# are the legs of of the right triangle. Since #AB=4# and #BP=3#, the length of the hypotenuse #AP=sqrt(4^2+3^2)=5#.

Now consider triangle #APQ#. Two of its sides are of length #sqrt(5)# and #sqrt(20)#. Note that if we add squares the length of both of these sides, we get the square of length of the third side

#(sqrt(5))^2+(sqrt(20))^2=5^2#

which means that triangle #APQ# **must** be a right triangle. We have already proven that the ratio of the length of its longer leg to its shorter leg is 2. The same is true for right triangle #CPQ#, so #APQ# and #CPQ# must be similar (by the **S** ide- **A** ngle- **S** ide Theorem).