The two squares ABPQ and ADRS drawn on two sides AB and AD of the parallelogram ABCD respectively are outside of ABCD. Let us prove that, PRC is an isosceles triangle.

In figure, ABCD is a parallelogram, and squares ABPQ and ADRS drawn on two sides AB and AD.

To prove: PRC is an isosceles triangle i.e. CP = CR

Now,

AB = CD [Opposite sides of parallelogram are equal]

⇒ BP = CD [∵ ABPQ is square, AB = BP = PQ = AQ] ……[1]

Similarly,

DR = BC ……[2]

Also,

∠ABP = ∠ADR = 90° [Angles in squares]

and ∠ABC = ∠ADC [opposite angles of a parallelogram are equal]

⇒ ∠ABC + ∠ABP = ∠ADR + ∠ADC

⇒ ∠CBP = ∠CDR ……[3]

Now, In ΔCBP and ΔCDR

BP = CD [From 1]

DR = BC [From 2]

∠CBP = ∠CDR [From 3]

ΔCBP ≅ ΔCDR [By SAS congruency criterion]

CP = CR [Corresponding parts of congruent triangles are equal]

Hence Proved!