Construct a triangle ABC in which angle ABC = 75°, AB = 5 cm and BC = 6.4 cm. Draw perpendicular bisector of side BC and also the bisector of angle ACB. If these bisectors intersect each other at point P; prove that P is equidistant from B and C; and also from AC and BC.

Steps of Construction:

1. Draw a line segment BC = 6.4 cm

2. At B, draw a ray BX making an angle of 75° and cut off BA = 5 cm.

3. Join AC. ∆ABC is the required triangle.

4. Draw the perpendicular bisector of BC, intersecting BC at Q.

5. Draw CX, the angle bisector of angle ACB which intersects the perpendicular bisector of BC at P.

6. Join PB and draw PL ⊥ AC.

In ∆PBQ and ∆PCQ,

⇒ PQ = PQ [Common]

⇒ ∠PQB = ∠PQC [Each = 90°]

⇒ BQ = QC [As PQ is the perpendicular bisector of BC]

∴ ∆PBQ ≅ ∆PCQ [By SAS axiom]

∴ PB = PC [By C.P.C.T.]

∴ P is equidistant from B and C.

Also,

In ∆PQC and ∆PLC,

⇒ ∠PQC = ∠PLC [Each 90°]

⇒ ∠PCQ = ∠PCL [As CX is angle bisector of ∠ACB]

⇒ PC = PC [Common]

∴ ∆PQC ≅ ∆PLC by AAS axiom.

∴ PQ = PL [By C.P.C.T.]

∴ P is equidistant from AC and BC.

Hence, proved that P is equidistant from B and C and also from AC and BC.