Mathematics
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.
Locus
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Answer
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. Then, ∆ABC is the required triangle.
Draw LM, perpendicular bisector of BC.
Let LM intersect AC at D and BC at E.
Join DB.
In ∆DBE and ∆DCE,
⇒ BE = EC [LM is the perpendicular bisector of BC]
⇒ ∠DEB = ∠DEC [Each 90°]
⇒ DE = DE [Common]
∴ ∆DBE ≅ ∆DCE [By SAS]
∴ DB = DC [By C.P.C.T.]
Hence, proved that D is equidistant from B and C.
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