\( \newcommand{\degree}{^{\circ}} \)

Proof

First, we are going to prove the following lemma:

Lemma: Any triangle is isosceles w.r.t. arbitrary two sides (i.e. any two sides are of equal length).
Proof: \( \square \) Let the triangle be \( ABC \). We’ll prove that \( AC = BC \). Here is a figure of the triangle with a few additional constructions, which we’ll define bellow:

Figure of \( \triangle ABC \)

Constructions: Let \( M_C \) be the midpoint of side \( AB \). Let \( l_C \) be the angle bisector of \( \angle ACB \). Let \( h_C \) be the perpendicular bisector of side \( AB \), which means that \( M_C \in h_C \). Let \( H_A \) be the projection of \( D \) on \( AC \) and, similarly, let \( H_B \) be the projection of \( D \) on \( BC \).

Justification: By SAS, \( DM_C \) = \( DM_C \), \( \angle AM_CD = 90\degree = \angle BM_CD \) and \( AM_C = BM_C \) we have that \( \triangle AM_CD \cong \triangle BM_CD \). Angle \[ \begin{align} \angle H_ADC = &\; 180\degree - & \angle DH_AC - & \angle H_ACD \\ = &\; 180\degree - & 90\degree - & \angle H_ACD \\ = &\; 180\degree - & \angle DH_BC - & \angle H_BCD = \angle H_BDC \end{align} \] By ASA, \( \angle H_ACD = \angle H_BCD \), \( CD = CD \) and \( \angle H_ADC = \angle H_BDC \) we have that \( \triangle CH_AD \cong \triangle CH_BD \). Since \( \triangle CH_AD \cong \triangle CH_BD \) then \( H_AD = H_BD\) and \( H_AC = H_BC \) . Since \( \triangle AM_CD \cong \triangle BM_CD \) then \( AD = BD \). Triangles \( \triangle AH_AD \) and \( \triangle BH_BD \) are right triangles, since \( \angle AH_AD = 90\degree = \angle BH_BD \). So, by Pythagorean theorem \[ AH_A = \sqrt{ AD^2 - H_AD^2 } = \sqrt{ BD^2 - H_BD^2 } = BH_B \]. Thus, \( AC = AH_A + H_AC = BH_B + H_BC = BC \) \( \blacksquare \)

Now, we use our lemma to prove the following theorem:

Theorem: Any triangle is equilateral (i.e. all three sides are of equal length).
Proof: \( \square \) Let \( \triangle ABC \) be the triangle. By using the proven lemma on \( \triangle ABC \), \( AC = BC \). Applying the same lemma on \( \triangle ACB \), \( AB = CB\). To conclude, \( AC = BC = AB \) \( \blacksquare \)

Cover photo by Tim Mossholder on Unsplash

Last modified on 2022-04-01