139. The Perpendicular Drop · Constructing a right angle from outside

Try to break it

Drag P closer to line AB or further away. The arcs and the intersection point E adjust to keep PQ exactly perpendicular to AB. Try to tilt PQ off 90° — you can't, that's the construction's promise.

How you build it

Construct a perpendicular from an external point.

• Draw line AB across the canvas. This is the line you will drop a perpendicular onto.
• Mark point P anywhere above line AB. P is the external point from which you will drop the perpendicular.
• Place the compass on P and draw an arc that crosses AB at two points.
• Mark point C where the arc crosses AB on the left.
• Mark point D where the arc crosses AB on the right.
• Open the compass to a new radius — wider than half of CD. Place it on C and swing an arc below line AB.
• Keep the same radius. Place the compass on D and swing an arc below AB that crosses the arc from C.
• Mark point E where the two arcs intersect below line AB.
• Use the straightedge to draw a line from P through E. It crosses AB at Q, and PQ is perpendicular to AB.

The proof, step by step

Prove that PQ is perpendicular to AB from the external point.

• PC = PD (radii of the same arc drawn from P)
• EC = ED (radii of equal arcs drawn from C and D)
• Therefore, P and E both lie on the perpendicular bisector of segment CD.
• Since CD lies on AB, line PE is perpendicular to AB at Q.

Worked example

In the given figure, PQ is drawn perpendicular to line AB from an external point P using the arc construction method. If PC = 15 cm and CD = 18 cm, what is the length of PQ?

Q is the midpoint of CD, so CQ = 9 cm. In right triangle PQC, PQ² + CQ² = PC². PQ² + 81 = 225, so PQ² = 144, giving PQ = 12 cm.

• 12 cm — correct
• 10 cm
• 9 cm
• 15 cm