Question

IN triangle ABC the altitude AD , BE and CF are equal . Prove that traingle ABC is an equilateral triangle.

Answer 1

Given : AD = BE = CF and angles A = B = C= 90°

To prove : Triangle ABC is equialteral.

Proof : In triangle ABE and ACF

Angle AEB = AFC ( 90° each )

BE = CF ( Given )

Angle A = A ( Common )

Hence Triangle ABE ~ ACF by AAS congruency

AB = AC ( c.p.c.t )

In triangle AOE and EOC

OE = OE ( Common )

Angle E = E ( 90° each )

AO = OC [ :. AD = FC, their halfs OA = OC ]

Hence triangle AOE ~ EOC by RHS congruency

AE = EC ( c.p.c.t )

In triangle ABE and BCE

Angle E = E ( 90° each )

BE = BE ( common )

AE = EC ( proved above )

Hence, triangle ABE ~ BCE by SAS congruency.

AB = BC ( c.p.c.t )

As AB = AC and AB = BC, so AC = BC.

Hence, AB = BC = CA.

HENCE PROVED.

Answer 2

Answer:

In right triangles BEC and BFC, we have

Hypotenuse BC = Hypotenuse BC

BE = CF (given)

So, by Rhs-criterion of congruence, we have

$\implies$∆BCE $\cong$ ∆BFC

$\implies$ $\angle$B=$\angle$C (CPCT)

$\implies$ AC = AB ....(1) ( $\therefore$ Sides opposite to equal angles are equal)

_________________________

Similarly, ∆ABD $\cong$ ∆ABE

$\implies$$\angle$B=$\angle$C

$\implies$ AC = BC......(2) ( $\therefore$ Sides opposite to equal angles are equal)

From (1) and (2), we have

AB = BC = AC

Hence, ∆ABC is an equilateral triangle.