Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Lemma

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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

if and only if:

$\angle ADG = 15 \degrees \text { or } \angle CDH = 15 \degrees$ (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
\(\ds \) \(=\) \(\ds DG^2 - AD^2\)
\(\ds \) \(=\) \(\ds AG^2\)
\(\ds \leadsto \ \ \) \(\ds CH\) \(=\) \(\ds AG\)

So by Triangle Side-Side-Side Equality:

$\triangle GAD = \triangle CDH$

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.

From Internal Angle of Equilateral Triangle:

$\angle GDH = 60 \degrees$

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:

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

$\blacksquare$