Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Construction 4

Construction

 * Inscribing-equilateral-triangle-inside-square-4.png

By Construction of Equilateral Triangle, let an equilateral triangle $\triangle ABN$ be constructed on $AB$ such that $N$ is inside $\Box ABCD$.

Let $DN$ be produced to cut $BC$ at $H$.

Construct $H$ on $BC$ such that $DH = DG$.

Then $DGH$ is the required equilateral triangle.

Proof
First a lemma:

Lemma
Because $\triangle ABN$ is equilateral:
 * $AB = AN$
 * $\angle BAN = 60 \degrees$

Thus $\triangle ADN$ is isosceles with apex at $A$.

Then $\angle DAN = 90 \degrees - 60 \degrees = 30 \degrees$.

From Sum of Angles of Triangle equals Two Right Angles:
 * $\angle ADN + \angle AND = 180 \degrees - 30 \degrees = 150 \degrees$

From Isosceles Triangle has Two Equal Angles:
 * $\angle ADN = \angle AND = \dfrac {150 \degrees} 2 = 75 \degrees$

Thus:
 * $\angle CDH = 15 \degrees$

The result follows from the lemma.