Suppose ABC is a triangle in which BE and CF are respectively the perpendiculars to the sides AC and AB. If BE = CF, prove that triangle ABC is isosceles.

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Solution

Given: BE = CF and ∠BEA = ∠CFA = 90°

To Prove: AB = AC

Proof:

In ΔABE and ΔFCA:

BE = CF (Given)

∠BEA = ∠CFA (Given)

∠A = ∠A (Common)

∴ ΔABE ≅ ΔFCA (AAS congruency)

⇒ AB = AC (Corresponding parts of congruent triangles)

Thus, the given triangle is isosceles.

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Suppose ABC is a triangle in which BE and CF are respectively the perpendiculars to the sides AC and AB. If BE = CF, prove that triangle ABC is isosceles.

Given: BE = CF and ∠BEA = ∠CFA = 90° To Prove: AB = AC Proof: In ΔABE and ΔFCA: BE = CF (Given) ∠BEA = ∠CFA (Given) ∠A = ∠A (Common) ∴ ΔABE ≅ ΔFCA (AAS congruency) ⇒ AB = AC (Corresponding parts of congruent triangles) Thus, the given triangle is isosceles.

Given: BE = CF and ∠BEA = ∠CFA = 90°

To Prove: AB = AC

Proof:

In ΔABE and ΔFCA:

BE = CF (Given)

∠BEA = ∠CFA (Given)

∠A = ∠A (Common)

∴ ΔABE ≅ ΔFCA (AAS congruency)

⇒ AB = AC (Corresponding parts of congruent triangles)

Thus, the given triangle is isosceles.