# Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Construction 2

## 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 $\triangle DAI$ be constructed on $AD$ such that $I$ is inside $\Box ABCD$.

Bisect $\angle ADI$ and bisect it again towards $AD$ 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
$\angle ADG = 15 \degrees \text { or } \angle CDH = 15 \degrees$ (and in fact both are the case).

$\Box$

Because $\triangle DAI$ is equilateral, we have that:

 $\ds \angle ADI$ $=$ $\ds 60 \degrees$ $\ds \leadsto \ \$ $\ds \angle ADG$ $=$ $\ds 15 \degrees$ having bisected it twice

The result follows from the lemma.

$\blacksquare$