Geometry Question on Quadrilaterals [Triangle and Square]

Question

Q) ABCD is a square. Prove that Triangle ABP is equilateral

My attempt:as PDC is isosceles, PD = PC, AD = BC and ∠ADP = ∠BCP
therefore, Triangles ADP and BPC are congruent
so BP=AP.. now i don't get how i can get AB=AP=PB

Answer 1

This is a case where it's easiest to work backwards. Construct the equilateral triangle $AP'B$. Then triangles $\triangle ADP'$ and $\triangle BP'C$ are obviously isosceles and congruent, with the base angles $\frac{1}{2}(180^\circ-30^\circ)=75^\circ$ each. This leaves $\widehat{P'AB} = \widehat{P'BA}=90^\circ - 75^\circ = 15^\circ$.

It follows that the lines $P'A \equiv PA$ and $P'B \equiv PB$ are in fact identical, since they make the same angles with $AB$. It then follows that their unique intersection must be $P' \equiv P$, so in the end triangle $\triangle APB \equiv \triangle AP'B$ is equilateral.