170. Diagonals of a Square · equal, bisecting, and perpendicular

Try to break it

Drag A around the circle. The square ABCD rotates with it, but the diagonals AC and BD always stay equal in length, bisect each other at O, and cross at 90°. Try to make any of those three properties fail — you can't. That's the full square diagonal theorem.

How you build it

Construct a square with diagonals.

• Mark point O — the centre of the circumscribed circle. All four vertices of the square will lie on this circle.
• Draw a circle with centre O using the circle tool.
• Mark a point A on the circumference of the circle.
• Construct a line through O and A, extending it to meet the circle again at C.
• Draw the perpendicular bisector of AC to find points B and D on the circle.
• Draw segment AB, the first side of the square.
• Draw segment BC, the second side of the square.
• Draw segment CD, the third side of the square.
• Draw segment DA to complete the square ABCD.

The proof, step by step

Prove that the diagonals of a square are equal and bisect each other at right angles.

• In ΔABC and ΔBAD: AB = AB (Common), AD = BC (Sides of a square), ∠ABC = ∠BAD = 90° ⇒ ΔABC ≅ ΔBAD (SAS) ⇒ AC = BD.
• In ΔAOB and ΔCOD: AB = DC (Sides), ∠OAB = ∠OCD (Alt. angles), ∠OBA = ∠ODC (Alt. angles) ⇒ ΔAOB ≅ ΔCOD (ASA) ⇒ OA = OC and OB = OD.
• In ΔAOB and ΔBOC: OA = OC (Proved), OB = OB (Common), AB = BC (Sides) ⇒ ΔAOB ≅ ΔBOC (SSS) ⇒ ∠AOB = ∠BOC.
• ∠AOB + ∠BOC = 180° (Linear pair on AC) ⇒ 2∠AOB = 180° ⇒ ∠AOB = 90°. Hence, diagonals bisect at 90°.

Worked example

In a square ABCD, the diagonals AC and BD intersect at O. If the length of diagonal AC is 16 cm, what is the length of segment OA?

Diagonals of a square bisect each other. Therefore, OA = AC / 2 = 16 / 2 = 8 cm.

• 4 cm
• 6 cm
• 8 cm — correct
• 12 cm