In the parallelogram ABCD, ∠BAD is an obtuse angle; the two equilateral triangles ABP and ADQ are drawn on the two sides AB & AD outside of it. Let us prove that, CPQ is an equilateral triangle.

In figure, ABCD is a parallelogram, and equilaterals ABP and ADQ drawn on two sides AB and AD.

To prove: CPQ is an equilateral triangle i.e. CP = CQ = PQ

Now,

AB = CD [Opposite sides of parallelogram are equal]

⇒ BP = CD [∵ ABP is an equilateral triangle, AB = BP = PQ ] ……[1]

Similarly,

DQ = BC ……[2]

Also,

∠ABP = ∠ADQ = 60° [Angles in equilateral triangles]

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

⇒ ∠ABC + ∠ABP = ∠ADQ + ∠ADC

⇒ ∠CBP = ∠CDQ ……[3]

Now, In ΔCBP and ΔCDQ

BP = CD [From 1]

DQ = BC [From 2]

∠CBP = ∠CDQ [From 3]

ΔCBP ≅ ΔCDQ [By SAS congruency criterion]

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

Now,

∠PAQ + ∠DAQ + ∠BAD + ∠PAB = 360°

⇒ ∠PAQ + 60° + ∠BAD + 60° = 360°

⇒ ∠PAQ + ∠BAD = 240°

Also,

∠BAD + ∠ADC = 180° [Interior angles on the same side of a transversal]

⇒ ∠PAQ + 180° - ∠ADC = 240°

⇒ ∠PAQ = 60° + ∠ADC

⇒ ∠PAQ = ∠QDA + ∠ADC [As, ∠QDA = 60°]

⇒ ∠PAQ = ∠QDC ……[4]

And

AP = AB [Sides of equilateral triangle]

AB = CD [Opposite sides of an equilateral triangle are equal]

⇒ AP = CD ……[5]

Now, In ΔAPQ and ΔDCQ

AP = CD [From 5]

AQ = QD [Sides of equilateral triangle]

∠PAQ = ∠QDC [From 4]

ΔAPQ ≅ ΔDCQ [By SAS congruency criterion]

PQ = CQ [Corresponding parts of congruent triangles are equal]

∴ CP = CQ = PQ

Hence Proved!