# Minimum steps to prove four points form a square.

I have points $$A, B, C, D$$ and I need to show that $$ABCD$$ is a square. I tried this by showing all four sides have equal length and that two angles sharing a side are $$90^{\circ}$$ degrees. Is there a more concise way to prove the points form a square?

• Diagonals of a square are equal. All sides are equal. Diagonals bisect each other at $90^{\circ}$. – Paras Khosla Mar 26 at 11:02
• One point finer than yours is that if the shared side is BC you only have to show AB BC and CD are equal (unless the shape can be self intersecting). – Paul Childs Mar 26 at 11:39

Prove that $$AC\perp BD$$, $$AC=BD$$ and $$AC$$ and $$BD$$ have the same midpoint.