Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Lemma

Lemma
Let $\Box ABCD$ be a square.

Let $\triangle DGH$ be an isosceles triangle inscribed within $\Box ABCD$ such that the apex $D$ of $\triangle DGH$ coincides with vertex $D$ of $\Box ABCD$.


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

Then:
 * $\triangle DGH$ is equilateral triangle


 * $\angle ADG = 15 \degrees \text { or } \angle CDH = 15 \degrees$ (and in fact both are the case).
 * $\angle ADG = 15 \degrees \text { or } \angle CDH = 15 \degrees$ (and in fact both are the case).

Proof
First note that $\triangle DGH$ is isosceles.

First we note that:

Then:

So by Triangle Side-Side-Side Equality:
 * $\triangle ADG = \triangle CDH$

and in particular:
 * $\angle ADG = \angle CDH$

Necessary Condition
Let $\angle ADG = \angle CDH = 15 \degrees$.

Then:
 * $\angle GDH = 90 \degrees - 2 \times 15 \degrees = 60 \degrees$

We have that $\triangle DGH$ is isosceles.

Hence from Isosceles Triangle has Two Equal Angles:
 * $\angle DGH = \angle DHG$

From Sum of Angles of Triangle equals Two Right Angles it follows that:
 * $\angle DGH + \angle DHG = 180 \degrees - 60 \degrees = 120 \degrees$

from which:
 * $\angle DGH = \angle DHG = 60 \degrees$

Hence all the vertices of $\triangle DGH$ equal $60 \degrees$.

It follows from Equiangular Triangle is Equilateral that $\triangle DGH$ is equilateral.

Sufficient Condition
Let $\triangle DGH$ be equilateral.

From Internal Angle of Equilateral Triangle:
 * $\angle GDH = 60 \degrees$

Because $\Box ABCD$ is a square:
 * $\angle ADC = 90 \degrees$

Thus:

But we have that:
 * $\angle ADG = \angle CDH$

and so:
 * $\angle ADG = \angle CDH = 15 \degrees$