Question

P is a point equidistant from two lines m and n intersecting at a point M. Show that the line MP bisects the angle between the lines m and n.

Answer 1

Given: P is a point equidistant from two lines m and n which intersect at M.

To Prove: The line MP bisects the angle between lines m and n; i.e., ∠AMP = ∠PMB, where A ∈ m and B ∈ n.

Proof:

1. Let A be the foot of the perpendicular from P to line m and B be the foot of the perpendicular from P to line n.

Then PA ⟂ m and PB ⟂ n by definition of perpendicular foot.

2. Because P is equidistant from m and n, the perpendicular distances are equal: PA = PB   ...(Given)

3. Consider right triangles ΔPAM and ΔPBM.

∠PAM = 90° (since PA ⟂ m) and ∠PBN = 90° (since PB ⟂ n), so both triangles are right-angled at A and B, respectively.

PM is common to both triangles.

PA = PB   ...(From step 2)

4. By the RHS (right angle–hypotenuse–side) congruence criterion, ΔPAM ≅ ΔPBM (right angle, hypotenuse PM common and corresponding leg PA = PB).

5. From the congruence, corresponding angles are equal.

Hence, ∠AMP = ∠PMB.

MP bisects the angle between lines m and n, as required.