Mathematics
In the adjoining figure, AD, BE and CF are altitudes of △ABC. If AD = BE = CF, prove that ABC is an equilateral triangle.
Triangles
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Answer
AD, BE and CF are altitudes of △ABC and
AD = BE = CF
Considering △BEC and △BFC,
Hypotenuse BC = BC (Common)
Side BE = CF (Given)
As altitudes are perpendicular to the sides,
∠CFB = ∠CEB = 90°
Hence, △BEC ≅ △BFC by RHS axiom.
We know that corresponding parts of congruent triangle are equal,
∴ ∠C = ∠B
⇒ AB = AC (Sides opposite to equal angles) ………(i)
Considering △CFA and △ADC,
AD = CF (Given)
∠ADC = ∠CFA = 90° (As altitudes are perpendicular to sides)
AC = AC (Common)
Hence, △CFA ≅ △ADC by RHS axiom.
We know that corresponding parts of congruent triangle are equal,
∴ ∠A = ∠C
⇒ AB = BC (Sides opposite to equal angles) ………(ii)
From (i) and (ii)
AB = BC = AC
△ABC is an equilateral triangle.
Hence, proved that △ABC is an equilateral triangle.
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