Equilateral Triangle Inside a Square – Detailed Proof

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:

\[ \tan(75^\circ) = \tan(30^\circ + 45^\circ) = \frac{\tan 30^\circ + \tan 45^\circ}{1 - \tan 30^\circ \tan 45^\circ} \]

Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \) and \( \tan 45^\circ = 1 \), we obtain:

\[ \tan(75^\circ) = 2 + \sqrt{3} \]

Substitute and solve:

\[ 2 = \frac{1}{\tan(\angle MCM')} + \frac{1}{2 + \sqrt{3}} \]

Simplifying the constant term gives \( \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3} \), so:

\[ \frac{1}{\tan(\angle MCM')} = 2 - (2 - \sqrt{3}) = \sqrt{3} \]

After taking the reciprocal:

\[ \tan(\angle MCM') = \frac{1}{\sqrt{3}} \]

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: \(\triangle DMC\) is equilateral.