Regular triangles BCP and CDL are constructed externally on the sides BC and CD

Regular triangles BCP and CDL are constructed externally on the sides BC and CD of the parallelogram ABCD. Prove that triangle APL is regular.

Triangles ABP and ADL are equal (on 2 sides and the angle between them):

AB = DC (parallelogram), DC = DL (since equilateral, regular triangle).

Similarly BP = AD;

Angles ABP = ADL = 360 – 60 – ABC = 360 – 60 – ADC (ABC = ADC since parallelogram) => AP = AL

In triangle PCL, the two sides are the same as in triangle ADL (reasoning is similar).

Angle PCL = 60 + 60 + BCD = 120 + 180 – ADC = 300 – ADC = ADL => and this triangle is equal to the two discussed above => AP = AL = PL.

Q.E.D.