Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Construction 3

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 $\triangle CDN$ be constructed on $CD$ such that $N$ is inside $\Box ABCD$.

Bisect $\angle ADN$ to cut $AB$ at $G$.

Construct $H$ on $BC$ 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 $\triangle CDN$ is equilateral, we have that:

\(\ds \angle DNC\)

\(=\)

\(\ds 60 \degrees\)

\(\ds \leadsto \ \ \)

\(\ds \angle ADN\)

\(=\)

\(\ds 90 \degrees - 60 \degrees\)

\(\ds \)

\(=\)

\(\ds 30 \degrees\)

\(\ds \leadsto \ \ \)

\(\ds \angle ADG\)

\(=\)

\(\ds 15 \degrees\)

having bisected $\angle ADN$

The result follows from the lemma.

$\blacksquare$