Instruction: Let |AB|=1. Construct a point C on the ray AB such that the length of AC is equal to √3.
Goal: 3L 3E
- Perpendicular Bisector
- Angle Bisector
Previous level: 4.6
Next level: 4.8
3L 3E solution
- Construct the circle with center B and radius AB, intersecting the ray at D
- Construct the circle with center D and radius BD, intersecting circle B at E
- Construct the circle with center A and radius AE, intersecting the ray at C
- Point at C
- AB = 1 = BD, DE = 1
- AD = 2
- AED = 90o based on Thales Theorem
Using the 30-60-90 special triangle, we can deduce that side AE = . AE and AC are both radii of the circle centered at A.