Prove that a sufficient condition because the triangle $ΔABC$ is equilateral is the following:
The height $CH$ on the side $AB$ divides $AB$ into two equal parts, and $AC = 2AH.$
It is necessary?
Prove that it is also a sufficient condition because
$(a) ∠AHC = ∠BHC$ and because
$(b) ∠AHC = \dfracπ2$.
$\triangle ABC$ is such that, if the height of $AB$ is $CH$, then:
$AH=HB$, $AC=2AH$.
We prove that $\triangle ABC$ is equilateral.
$AH=HB\Rightarrow AB=2AH\Rightarrow AB=AC$ (two equal sides)
$CH\perp AB\Rightarrow In\hspace{2mm} \triangle CHA, \cos (HAC)=\dfrac{AH}{AC}=\dfrac{AH}{2AH}=\dfrac{1}{2}\Rightarrow HAC=A=60º$
By the theorem of cosinus (in $\triangle ABC$):
$BC^{2}=AC^{2}+AB^{2}-2AB\cdot AC\cdot \cos (60º)\Rightarrow BC^{2}=AC^{2}+AC^{2}-2AC^{2}\cdot \dfrac{1}{2}=AC^{2}+0\Rightarrow BC^{2}=AC^{2}\Rightarrow BC=AC=AB\Rightarrow \triangle ABC \hspace{2mm} equilateral$
It is evident that: $\triangle ABC$ equilateral $\Rightarrow$ $AH=HB$, $AC=2AH$.