# 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.

$\blacksquare$

## Sources

- 1986: J.L. Berggren:
*Episodes in the Mathematics of Medieval Islam* - 1992: David Wells:
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