Inscribing Equilateral Triangle inside Square with a Coincident Vertex/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).
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 Congruence:
- $\triangle ADG = \triangle CDH$
and in particular:
- $\angle ADG = \angle CDH$
Necessary Condition
Let $\angle ADG = \angle CDH = 15 \degrees$.
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:
\(\ds \angle ADG + \angle GDH + \angle CDH\) | \(=\) | \(\ds 90 \degrees\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \angle ADG + \angle CDH\) | \(=\) | \(\ds 90 \degrees - \angle GDH\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 90 \degrees - 60 \degrees\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 30 \degrees\) |
But we have that:
- $\angle ADG = \angle CDH$
and so:
- $\angle ADG = \angle CDH = 15 \degrees$
$\blacksquare$