Question

AD, BE and CF are the altitudes of ∆ABC are equal. Prove that ∆ABC is an equilateral
triangle.​

Answer 1

Answer:

Step-by-step explanation:

• In triangle ABE and triangle ACF

BE=CF (Given)

∠A=∠A (Common)

∠AEB=∠AFC=90°

By AAS congruent rule ABE ≅ ACF

AB=AC (CPCT)

Similarly

BCF ≅ ABD

AB=BC

AC=BC

AB=AC

AB = BC = AC..... Proved

©© The triangle is a equaliteral triangle ©©

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Answer 2

Answer:

GIVEN THAT -

AD ,BE, CF are the altitude of triangle ABC.

AD ,BE, CF are equal.

TO PROVE -

The Triangle ABC is equilateral Triangle.

So,

In triangle ABE and triangle ACF.

As given, BE = CF

⇒ ∠A = ∠A [ Common ]

So,

⇒ ∠AEB = ∠AFC = 90°

Triangle DAC is congruent to Triangle CBE.

We know that -

The AAS property,

Statement -

Two triangle are congruent if any two pair of angles and one pair of corresponding sides are equal.

SO,

It implies,

⇒ ABE is congruent to ACF.

THEREFORE,

⇒ AB = AC

NOW,

BCF ≅ ABD

⇒ AB = BC

⇒ AC = BC

⇒ AB = AC

⇒ AB = BC = AC