321. The Bisector's Balance · equidistant from intersecting lines
The perpendicular distances from P to both lines remain exactly equal as P moves along the bisector.
What this lesson covers
Try to break it
Drag P along the angle bisector. The perpendicular distances PM and PN to the two sides always come out equal. Try to slide P off the bisector (in your head) and the equality breaks instantly — the bisector is exactly the locus of points equidistant from both sides.
How you build it
Construct the bisector of angle AOC with compass arcs.
- Draw the first line AB across the canvas.
- Draw the second line CD so it crosses AB. Their intersection is the vertex O of angle AOC.
- Put the compass point on O and draw an arc that cuts line AB at M and line CD at N.
- Put the compass on M and draw an arc in the interior of angle AOC.
- Keep the same radius, put the compass on N and draw an arc. It meets the arc from M at point X.
- Draw the line through O and X. OX is the bisector of angle AOC — it splits the angle into two equal halves.
- Mark a point P anywhere on the bisector OX.
- From P, drop a perpendicular to line AB. Its length is the distance from P to AB.
- From P, drop a perpendicular to line CD. It comes out the SAME length — P on the bisector is equidistant from both lines.
The proof, step by step
Prove that the bisector of two intersecting lines is the locus of points equidistant from both.
- Draw perpendiculars PM and PN from P to lines AB and CD respectively.
- In ΔOPM and ΔOPN: ∠OMP = ∠ONP = 90°, OP is common, and ∠POM = ∠PON (since OP bisects ∠AOC).
- By AAS congruence criterion, ΔOPM ≅ ΔOPN.
- Therefore, PM = PN. Hence, P is equidistant from AB and CD.
Worked example
In the figure, lines AB and CD intersect at O. P is a point on the bisector of ∠AOC. PM ⊥ AB and PN ⊥ CD. If PM = 5 cm, what is the length of PN?
Since P lies on the bisector of ∠AOC, it is equidistant from the arms AB and CD. Therefore, PN = PM = 5 cm.
- 2.5 cm
- 5 cm — correct
- 7.5 cm
- 10 cm