Texas HS Geometry Question of the Day

$AB \cong AC$ because $$d_{AB}$$=d_{AC}$=4$\sqrt{2}$$.

$BC \cong AC$ because $$d_{BC}$=4$\sqrt{2}$$.

No two sides are congruent.

$AB \cong BC$ because $$d_{AB}$=$\sqrt{(5-(-3))^2+(-2-2)^2}$=$\sqrt{80}$=4$\sqrt{5}$$ and $$d_{BC}$=$\sqrt{(1-5)^2+(6-(-2))^2}$=$\sqrt{80}$=4$\sqrt{5}$$.