# 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 $\Box ABCD$.

Then:

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

## Proof

First note that $\triangle DGH$ is isosceles.

First we note that:

 $\ds CD$ $=$ $\ds AD$ as they are the sides of a square $\ds DH$ $=$ $\ds DG$ as $\triangle DGH$ is isosceles $\ds \angle DCH$ $=$ $\ds \angle DAG = 90 \degrees$ as $\Box ABCD$ is a square

Then:

 $\ds CH^2$ $=$ $\ds DH^2 - CD^2$ Pythagoras's Theorem $\ds$ $=$ $\ds DG^2 - AD^2$ $\ds$ $=$ $\ds AG^2$ $\ds \leadsto \ \$ $\ds CH$ $=$ $\ds AG$
$\triangle GAD = \triangle CDH$

and in particular:

$\angle GAD = \angle CDH$

### Necessary Condition

Let $\angle GAD = \angle CDH = 15 \degrees$.

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.

$\Box$

### Sufficient Condition

Let $\triangle DGH$ be equilateral.

$\angle GDH = 60 \degrees$

Because $\Box ABCD$ is a square:

$\angle ADC = 90 \degrees$

Thus:

 $\ds \angle GAD + \angle GDH + \angle CDH$ $=$ $\ds 90 \degrees$ $\ds \leadsto \ \$ $\ds \angle GAD + \angle CDH$ $=$ $\ds 90 \degrees - \angle GDH$ $\ds$ $=$ $\ds 90 \degrees - 60 \degrees$ $\ds$ $=$ $\ds 30 \degrees$

But we have that:

$\angle GAD = \angle CDH$

and so:

$\angle GAD = \angle CDH = 15 \degrees$

$\blacksquare$