Question

In an equilateral triangle ABC, the midpoints of the sides are labelled D, E and F. Prove that triangle DEF is an equilateral triangle.

Answer 1

Answer:

Given, ABC is an equilateral triangle.D, E, F are mid points of the sides of the triangle.

Since, D is mid point of AB and E is mid point of AC, by mid point theorem,

DE= 1/2 AC........(1)

Since, F is mid point of BC and E is mid point of AC, by mid point theorem,

EF= 1/2 AB......(2)

And we know, BE= 1/2 AB and BF= 1/2 BC.........(3)

Now, from (1), (2) and (3)

Since, all the sides of equilateral triangle are equal,

DE=EF=BE=BF

Hence, BEFD is a rhombus.

OR.

Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively.

⇒ DE = 1 / 2 AB --- (i)

E and F are the mid - points of AC and AB respectively .

∴ EF = 1 / 2 BC --- (ii)

F and D are the mid - points of AB and BC respectively .

∴ FD = 1 / 2 AC --- (iii)

Now, △ABC is an equilateral triangle .

⇒ AB = BC = CA

⇒ 1 / 2 AB = 1 / 2 BC = 1 / 2 CA

⇒ DE = EF = FD [using (i) , (ii) , (iii) ]

Hence, DEF is an equilateral triangle

Answer 2

Answer:

proved

Step-by-step explanation:

Given: ABC is an equilateral triangle in which D, E, F are the midpoints of the sides AB, BC and AC respectively. RTP: DEF is an equilateral triangle Proof: AB = BC = AC (Given) In triangle ABC E and F are the mid point of AB and AC So, EF = BC/2 ------- (1) [∵Midpoint theorem] D and F are the midpoints of BC and AB So, DF = AC/2 [∵Midpoint theorem] ⇒ DF = BC/2 ------- (2) [∵AC = BC] D and E are the midpoints of BC and AC So, DE = AB/2 [∵Midpoint theorem] ⇒ DE = BC/2 ------- (3) [∵AB = BC] From (1), (2), and (3) ⇒EF = DF = DE ∴ DEF is an equilateral triangle .

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