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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.
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Solution
Given: In triangle ABC, AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm
Steps of Construction:
- Draw a line segment BC = 6.3 cm
- With centre B and radius 4.2 cm, draw an arc.
- With centre C and radius 5 cm, draw another arc which intersects the first arc at A.
- Join AB and AC. ΔABC is the required triangle.
- Again with centre B and C and radius greater than 12 BC, draw arcs which intersects each other at L and M.
- Join LM intersecting AC at D and BC at E.
- 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.
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