Question

Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C. 

Answer 1

Given: In triangle ABC, AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm 

 
Steps of Construction:

1. Draw a line segment BC = 6.3 cm 
2. With centre B and radius 4.2 cm, draw an arc.
3. With centre C and radius 5 cm, draw another arc which intersects the first arc at A.
4. Join AB and AC. ΔABC is the required triangle.
5. Again with centre B and C and radius greater than 12 BC, draw arcs which intersects each other at L and M.
6. Join LM intersecting AC at D and BC at E.
7. Join DB.

Proof: In ΔDBE and ΔDCE

BE = EC  ...(LM is bisector of BC)

∠DEB = ∠DEC  ...(Each = 90°)

DE = DE   ...(Common)

∴ By side angle side criterion of congruence, we have

ΔDBE ≅ ΔDCE   ...(SAS postulate)

The corresponding parts of the congruent triangle are congruent

∴ DB = DC  ...(C.P.C.T.)

Hence, D is equidistant from B and C.