Question

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. 

Answer 1

  
Steps of Construction:

1. Draw a line segment BC = 6.4 cm
2. At B, draw a ray BX making an angle of 75° with BC and cut off BA = 5 cm.
3. Join AC. ΔABC is the required triangle.
4. Draw the perpendicular bisector of BC.
5. Draw the angle bisector of angle ACB which intersects the perpendicular bisector of BC at P. 
6. Join PB and draw PL ⊥ AC.

Proof: In ΔPBQ and ΔPCQ

PQ = PQ   ...(Common)

∠AQB = ∠PQC  ...(Each = 90°)

BQ = QC   ...(PQ is the perpendicular bisector of BC)

∴ By side Angle side criterion of congruence

ΔPBQ ≅ ΔPCQ  ...(SAS Postulate)

The corresponding parts of the congruent triangle are congruent

∴ PB = PC  ...(C.P.C.T.)

Hence, P is equidistant from B and C.

∠PQC = ∠PLC  ...(Each = 90°)

∠PCQ = ∠PCL  ...(Given)

PC = PC  ...(Common)

Again in ΔPQC and ΔPLC

∴ By Angle – Angle side criterion of congruence,

ΔPQC ≅ ΔPLC  ...(AAS postulate)

The corresponding parts of the congruent triangles are congruent

∴ PQ = PL  ...(C.P.C.T.)

Hence, P is equidistant from AC and BC.