Question

2. Using a straightedge and compass, construct the perpendicular bisector of .

Answer:

Answer 1

Let the segment be AB.

Construction:
1. With compass centered at A, draw a circle of radius r, where r > 1/2·AB (choose any convenient radius larger than half the length of AB).
2. With the same radius r and the compass centered at B, draw a second circle.
3. Label the two intersection points of the circles P and Q.
4. Use the straightedge to draw the line PQ.

Result: Line PQ is the perpendicular bisector of AB. It meets AB at its midpoint M and is perpendicular to AB.

Justification:
- From the construction PA = PB (both are radius r) and QA = QB. Thus P and Q are equidistant from A and B.
- Any point equidistant from A and B lies on the perpendicular bisector of AB, so both P and Q lie on that bisector. The line through P and Q is therefore the perpendicular bisector.
- The intersection M = PQ ∩ AB is the midpoint of AB, and PQ ⟂ AB.