Geometry Practice Test 6

We know that
${\left(\mathit{a}+\mathit{b}+\mathit{c}\right)}^2$ = ${\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2$ + 2(ab + bc + ca)
a2 + b2 + ${\mathit{c}}^2$ = ab + bc + ca (given)
${\left(\mathit{a}+\mathit{b}+\mathit{c}\right)}^2$ = (ab + bc + ca)+ 2(ab + bc + ca)
${\left(\mathit{a}+\mathit{b}+\mathit{c}\right)}^2$ = 3 (ab + bc + ca)
${\left(3\mathit{a}\right)}^2$ = $3\left({\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2\right)$
${\left(3\mathit{a}\right)}^2$ = $3\left(3{\mathit{a}}^2\right)$
$9{\mathit{a}}^2$ = $9{\mathit{a}}^2$
It is possible only when the three sides of a triangle is equal.
So, the triangle is an equilateral triangle.