Answer
Proved below
Work Step by Step
Let, \[{{P}_{1}}=(0,0),\,{{P}_{2}}=(a,0),\,{{P}_{3}}=\left( \frac{a}{2},\frac{\sqrt{3}a}{2} \right)\]
So, the sides of triangle are ${{P}_{1}}{{P}_{2}},\,{{P}_{2}}{{P}_{3}},\,{{P}_{3}}{{P}_{1}}$
By the distance formula, the length of the side ${{P}_{1}}{{P}_{2}}$ is
$d({{P}_{1}},{{P}_{2}})=\sqrt{{{(a-0)}^{2}}+{{(0-0)}^{2}}}$
$d({{P}_{1}},{{P}_{2}})=\sqrt{{{(a)}^{2}}}$
$d({{P}_{1}},{{P}_{2}})=\left| a \right|$
By the distance formula, the length of the side ${{P}_{2}}{{P}_{3}}$is
$d({{P}_{2}},{{P}_{3}})=\sqrt{{{\left( \frac{a}{2}-a \right)}^{2}}+{{\left( \frac{\sqrt{3}a}{2}-0 \right)}^{2}}}$
$d({{P}_{2}},{{P}_{3}})=\sqrt{{{\left( -\frac{a}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}a}{2} \right)}^{2}}}$
$d({{P}_{2}},{{P}_{3}})=\sqrt{\frac{{{a}^{2}}}{4}+\frac{3{{a}^{2}}}{4}}$
$d({{P}_{2}},{{P}_{3}})=\sqrt{\frac{4{{a}^{2}}}{4}}$
\[d({{P}_{2}},{{P}_{3}})=\sqrt{{{a}^{2}}}\]
\[d({{P}_{2}},{{P}_{3}})=\left| a \right|\]
By the distance formula, the length of side ${{P}_{3}}{{P}_{1}}$ is
$d({{P}_{3}},{{P}_{1}})=\sqrt{{{\left( 0-\frac{a}{2} \right)}^{2}}+{{\left( 0-\frac{\sqrt{3}a}{2} \right)}^{2}}}$
$d({{P}_{3}},{{P}_{1}})=\sqrt{\frac{{{a}^{2}}}{4}+\frac{3{{a}^{2}}}{4}}$
$d({{P}_{3}},{{P}_{1}})=\sqrt{\frac{4{{a}^{2}}}{4}}$
\[d({{P}_{3}},{{P}_{1}})=\sqrt{{{a}^{2}}}\]
\[d({{P}_{3}},{{P}_{1}})=\left| a \right|\]
Therefore, the sides of the triangle are the same
Therefore, the points $(0,0),(a,0)$ and $\left( \frac{a}{2},\frac{\sqrt{3}a}{2} \right)$ are vertices of an equilateral triangle.
By the midpoint formula, the coordinate $A=(x,y)$ of the midpoint of the line joining two points ${{P}_{1}}=(0,0)$ and${{P}_{2}}=(a,0)$ is
$x=\frac{0+a}{2},\,y=\frac{0+0}{2}$
$x=\frac{a}{2},\,y=0$
$A$ is the midpoint of ${{P}_{1}}{{P}_{2}}$
Therefore, $A=\left( \frac{a}{2},0 \right)$
We find:
$x=\frac{\frac{a}{2}+a}{2},\,y=\frac{\frac{\sqrt{3}a}{2}+0}{2}$
$x=\frac{3a}{4},\,y=\frac{\sqrt{3}\,a}{4}$
$B$is the midpoint of ${{P}_{2}}{{P}_{3}}$
$x=\frac{\frac{a}{2}+0}{2},\,y=\frac{\frac{\sqrt{3}\,a}{2}+0}{2}$
$x=\frac{a}{4},\,y=\frac{\sqrt{3}\,a}{4}$
$C$ is the midpoint of ${{P}_{2}}{{P}_{3}}$
Therefore, $C=\left( \frac{a}{4},\,\frac{\sqrt{3}\,a}{4} \right)$
Consider the triangle $ABC$, where$A=\left( \frac{a}{2},0 \right)$, $B=\left( \frac{3a}{4},\,\frac{\sqrt{3}\,a}{4} \right)$and $C=\left( \frac{a}{4},\,\frac{\sqrt{3}\,a}{4} \right)$
Let us find the length of side $AB,BC,AC$
By the distance formula, the length of the side $AB$ is
$d(A,B)=\sqrt{{{\left( \frac{3a}{4}-\frac{a}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}a}{4}-0 \right)}^{2}}}$
$d(A,B)=\sqrt{{{\left( \frac{a}{4} \right)}^{2}}+{{\left( \frac{\sqrt{3}a}{4} \right)}^{2}}}$
$d(A,B)=\sqrt{\frac{{{a}^{2}}}{16}+\frac{3{{a}^{2}}}{16}}$
$d(A,B)=\sqrt{\frac{4{{a}^{2}}}{16}}$
$d(A,B)=\sqrt{\frac{{{a}^{2}}}{4}}$
$d(A,B)=\left| \frac{a}{2} \right|$
By the distance formula, the length of the side $BC$ is
$d(B,C)=\sqrt{{{\left( \frac{3a}{4}-\frac{a}{4} \right)}^{2}}+{{\left( \frac{\sqrt{3}a}{4}-\frac{\sqrt{3}a}{4} \right)}^{2}}}$
$d(B,C)=\sqrt{{{\left( \frac{2a}{4} \right)}^{2}}+0}$
$d(B,C)=\sqrt{\frac{4{{a}^{2}}}{16}}$
$d(B,C)=\sqrt{\frac{{{a}^{2}}}{4}}$
$d(B,C)=\left| \frac{a}{2} \right|$
By the distance formula, the length of the side $AC$ is
$d(A,C)=\sqrt{{{\left( \frac{a}{4}-\frac{a}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}a}{4}-0 \right)}^{2}}}$
$d(A,C)=\sqrt{{{\left( -\frac{a}{4} \right)}^{2}}+{{\left( \frac{\sqrt{3}a}{4} \right)}^{2}}}$
$d(A,C)=\sqrt{\frac{{{a}^{2}}}{16}+\frac{3{{a}^{2}}}{16}}$
$d(A,C)=\sqrt{\frac{4{{a}^{2}}}{16}}$
$d(A,C)=\sqrt{\frac{{{a}^{2}}}{4}}$
$d(A,C)=\left| \frac{a}{2} \right|$
Therefore, the length of sides $AB,BC,AC$ are the same
Therefore, the triangle $ABC$ is an equilateral triangle
