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

Construction for Inscribing Equilateral Triangle inside Square with a Coincident Vertex

Let $\Box ABCD$ be a square.

It is required that $\triangle DGH$ be an equilateral triangle inscribed within $\Box ABCD$ such that vertex $D$ of $\triangle DGH$ coincides with vertex $D$ of $\Box ABCD$.

Construction

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 $G$ on $AB$ such that $DH = DG$.

Then $DGH$ is the required equilateral triangle.

Proof

First a 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

if and only if:

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

$\Box$

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.

$\blacksquare$