Equilateral Triangle Inside a Square – Detailed Proof
In the figure below, \(ABCD\) is a square. We prove that triangle \(DMC\) is equilateral using congruent triangles and trigonometry.
Problem
Given square \(ABCD\), prove that triangle \(DMC\) is equilateral.
Solution
-
Triangle \(AMB\) is isosceles, so
\[
MA = MB
\]
Angles \(MAD\) and \(MBC\) are equal.
Thus triangles \(DAM\) and \(CBM\) have:
- Two equal sides
- The included equal angle
Therefore, the triangles are congruent (SAS).
Hence:
\[
DM = CM
\]
Triangle \(DCM\) is therefore isosceles.
-
Draw the perpendicular \(MM'\) to side \(CB\).
Consider right triangles \(MCM'\) and \(MBM'\).
Using tangent definitions:
\[
\tan(\angle MCM') = \frac{MM'}{CM'}
\]
\[
\tan(\angle MBM') = \frac{MM'}{BM'}
\]
-
Rewrite:
\[
CM' = \frac{MM'}{\tan(\angle MCM')}
\]
\[
BM' = \frac{MM'}{\tan(\angle MBM')}
\]
-
Since \( \angle MBM' = 75^\circ \), let the side of the square be \(x\).
Then:
\[
MM' = \frac{x}{2}
\]
\[
CM' + BM' = x
\]
-
Substitute:
\[
x = \frac{x/2}{\tan(\angle MCM')} + \frac{x/2}{\tan(75^\circ)}
\]
-
Divide by \(x\) and multiply by 2:
\[
2 = \frac{1}{\tan(\angle MCM')} + \frac{1}{\tan(75^\circ)}
\]
-
Compute \( \tan(75^\circ) \) using the tangent sum formula:
\[
\tan(75^\circ) = \tan(30^\circ + 45^\circ)
\]
\[
\tan(75^\circ)
=
\frac{\tan 30^\circ + \tan 45^\circ}
{1 - \tan 30^\circ \tan 45^\circ}
\]
Since
\[
\tan 30^\circ = \frac{1}{\sqrt{3}}, \quad
\tan 45^\circ = 1
\]
we obtain
\[
\tan(75^\circ) = 2 + \sqrt{3}
\]
-
Substitute:
\[
2 = \frac{1}{\tan(\angle MCM')} + \frac{1}{2 + \sqrt{3}}
\]
Solve:
\[
\frac{1}{\tan(\angle MCM')} = 2 - \frac{1}{2 + \sqrt{3}}
\]
After simplification:
\[
\tan(\angle MCM') = \frac{1}{\sqrt{3}}
\]
-
Therefore,
\[
\angle MCM' = 30^\circ
\]
-
Now compute:
\[
\angle DCM = 90^\circ - 30^\circ = 60^\circ
\]
-
Since triangle \(DCM\) is isosceles and one angle is \(60^\circ\), all angles are \(60^\circ\).
Therefore:
\[
\triangle DMC \text{ is equilateral.}
\]
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