Question

If the altitude drawn from the vertices of abc to the opposite sides are equal, prove that the triangle is equilateral.

Answer 1

If the altitude drawn from the vertices of abc to the opposite sides are equal, then the triangle is equilateral.

Solution:

To prove: If the altitude drawn from the vertices of abc to the opposite sides are equal, prove that the triangle is equilateral.

Consider a traingle ABC

The figure is attached below

AD and BE and CF is the altitudes drawn sides BC and AC and AB respectively

Given that altitude drawn from the vertices of abc to the opposite sides are equal

So AD = BE = CF

angles A = B = C= 90°

In triangle ABE and ACF

Angle AEB = AFC ( both are 90 degrees)

BE = CF ( Given that altitudes are equal)

Angle A = A ( Common in both triangle ABE and ACF)

The Angle Angle Side postulate states that if two angles and the non-included side of one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent.

Hence Triangle $A B E \cong A C F$ by AAS congruency

“Corresponding Parts of Congruent Triangles” states that if we take two or more triangles which are congruent to each other then the corresponding angles and the sides of the triangles are also congruent to each other i.e., their corresponding parts are equal to each other.

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.CT )

In triangle ABE and BCE

Angle E = E ( 90° each )

BE = BE ( common in both triangle ABE and BCE )

AE = EC ( proved above )  

The Side Angle Side postulate states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then these two triangles are congruent.

Hence, triangle $\mathrm{ABE} \cong \mathrm{BCE}$ by SAS congruency.

AB = BC ( C.P.CT )

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

Hence, AB = BC = CA

Since all three sides are equal, it is a equilateral triangle

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Answer 2

Answer:

GIVEN:

If the altitude drawn from the vertices of abc to the opposite sides are equal, then the opp. sides are equal

To prove: If the altitude drawn from the vertices of abc to the opposite sides are equal, prove that the triangle is equilateral.

Consider a traingle ABC

AD and BE and CF is the altitudes drawn sides BC and AC and AB respectively

Given that altitude drawn from the vertices of abc to the opposite sides are equal

So AD = BE = CF

angles ADC=BEC = CFB= 90°{ALTITUDES OF SAME TRIANGLES}

In triangle ABE and ACF

Angle AEB = AFC ( both are 90 degrees)

BE = CF ( Given that altitudes are equal)

Angle A = A ( Common in both triangle ABE and ACF)

Hence ABE≅ACF by AAS rule of congruence

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 rule of congruence

AE = EC ( C.P.CT )

In triangle ABE and BCE

Angle E = E ( 90° each )

BE = BE ( common in both triangle ABE and BCE )

AE = EC ( proved above )

Hence, triangle ABE≅BCE by SAS rule of congruence.

AB = BC ( C.P.CT )

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

Hence, AB = BC = CA

Since all three sides are equal, it is a equilateral triangle

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