136. The Perfect Right Angle · Constructing 90° with compass and straightedge

Try to break it

Drag A along the base. The construction scales with it, but OE always stands at a perfect 90° to OA. Try to tilt that right angle — the arcs refuse, locking it at exactly 90°.

How you build it

Construct a 90 degree angle.

• Draw a base ray from O going to the right — this is ray OA.
• With O as centre, draw an arc of any radius that crosses ray OA.
• Mark point A where the arc meets ray OA.
• With A as centre and the same radius, draw an arc that crosses the first arc.
• Mark point C where the arc from A crosses the first arc.
• With C as centre and the same radius, draw another arc that crosses the first arc.
• Mark point D where the arc from C meets the first arc.
• With C as centre and the same radius, draw an arc above the construction.
• With D as centre and the same radius, draw an arc crossing the arc from C.
• Mark point E where the arcs from C and D meet.
• Draw ray OE from O through E. Then ∠AOE = 90°.

The proof, step by step

Prove that the constructed ray OE is perpendicular to OA.

• OA = OC = OD (radii of the same initial arc).
• ΔOAC and ΔOCD are equilateral triangles, so ∠AOC = 60° and ∠COD = 60°.
• Arcs from C and D intersect at E, making CE = DE. Thus ΔOCE ≅ ΔODE by SSS.
• ∠COE = ∠DOE = 30°. Hence ∠AOE = ∠AOC + ∠COE = 60° + 30° = 90°.

Worked example

In the construction of a 90° angle at point O on line OA, if the radius used for arcs from C and D is equal to the initial radius, what is the measure of ∠COE?

Since OA=OC=OD=CE=DE, triangles OAC, OCD, OCE, and ODE are all equilateral or congruent isosceles. Specifically, ΔOCE ≅ ΔODE implies ∠COE = ∠DOE. Since ∠COD = 60°, ∠COE = 30°. Thus ∠AOE = 60° + 30° = 90°.

• 15°
• 30° — correct
• 45°
• 60°