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In the adjoining figure, AD, BE and CF are altitudes of △ABC. If AD = BE = CF, prove that ABC is an equilateral triangle.

In the adjoining figure, D is the midpoint of BC, DE and DF are perpendiculars to AB and AC respectively such that DE = DF. Prove that ABC is an isosceles triangle. Triangles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

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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