138. The Perpendicular Promise · always at right angles to the line

Try to break it

Drag P along AB. No matter where you put it on the line, PQ stays at a perfect 90°. Try to tilt it off — you can't, that's the theorem.

How you build it

Construct a perpendicular to a line at a point.

• Draw line AB across the canvas. This is the line you will construct a perpendicular to.
• Mark a point P anywhere on line AB. This is where you will raise the perpendicular.
• Place the compass point on P and draw an arc that crosses the line on both sides.
• Mark point C where the arc crosses the line on the left side of P.
• Mark point D where the arc crosses the line on the right side of P.
• Open the compass wider than before. Place the point on C and swing an arc above the line.
• Without changing the radius, place the compass on D and draw an arc that crosses the arc from C.
• Mark point Q where the two arcs intersect above the line.
• Use the straightedge to draw a line from P through Q. PQ is perpendicular to AB.

The proof, step by step

Prove that PQ is perpendicular to AB at the point on the line.

• By construction, PC = PD because they are radii of the same arc drawn from center P.
• With the same radius, arcs drawn from C and D intersect at Q, so CQ = DQ.
• PQ is common to both triangles PCQ and PDQ.
• Therefore, ΔPCQ ≅ ΔPDQ by SSS congruence rule.
• Hence, ∠QPC = ∠QPD. Since they form a linear pair (180°), each angle is 90°. Thus, PQ ⊥ AB.

Worked example

While constructing a perpendicular to a line segment AB at a point P on it, a student draws an arc from P cutting AB at C and D. If PC = 4 cm, what is the length of CD?

Since the arc is drawn with center P, PC and PD are radii of the same circle. Thus, PD = PC = 4 cm. The points C and D lie on opposite sides of P on the line, so CD = PC + PD = 4 + 4 = 8 cm.

• 2 cm
• 4 cm
• 8 cm — correct
• 16 cm