Sum
ABC is an equilateral triangle, AD and BE are perpendiculars to BC and AC respectively. Prove that:
(i) AD = BE
(ii) BD = CE
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Solution
In Δ ABC,
AB = BC = CA,
AD ⊥ BC, BE ⊥ AC.
Proof: In Δ ADC and Δ BEC
∠ADC = ∠BEC ...........(each 90°)
∠ACD = ∠BCE ..............(common)
and AC = BC ............(sides of an equilateral triangle)
∴ Δ ADC ≅ Δ BEC ...........(A.A.S. Axiom)
Hence (i) AD = BE ...........(c.p.c.t.)
and (ii) BD = CE ...........(c.p.c.t.)
Hence proved.
Concept: Extend Congruence to Simple Geometrical Shapes E.G. Triangles, Circles.
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