Prove that the figure determined by the points is an isosceles triangle. A(19,-1), B(13,7), C(5,1)

Prove that the figure determined by the points is an isosceles triangle. A(19,-1), B(13,7), C(5,1)
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Velsenw
Distance between line segment $\left(d\right)=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
Distance between A and B (AB) $=\sqrt{{\left(19-5\right)}^{2}+{\left(-1-1\right)}^{2}}$
$=\sqrt{{\left(14\right)}^{2}+{\left(-2\right)}^{2}}$
$=\sqrt{194+4}$
$=\sqrt{200}$
=14.142
AB=14.142
Distance between A and C (AC) $=\sqrt{{\left(19-13\right)}^{2}+{\left(-1-7\right)}^{2}}$
$=\sqrt{{\left(6\right)}^{2}+{\left(8\right)}^{2}}$
$=\sqrt{36+64}$
$=\sqrt{100}$
=10
AC=10
Distance between C and B (AC) $=\sqrt{{\left(13-5\right)}^{2}+{\left(7-1\right)}^{2}}$
$=\sqrt{{\left(8\right)}^{2}+{\left(6\right)}^{2}}$
$=\sqrt{64+36}$
$=\sqrt{100}$
=10
CB=10
Since AC=CB=10
Hence $\mathrm{△}ABC$ is an isosceles triangle.