Question

∆ABC is an equilateral triangle. BC is divided into three equal parts by points D and E. Is ∆ABD congruent to∆AEC?​

Answer 1

Answer:

Let ABC be the triangle and D, E and F be the mid-point of BC, CA and AB respectively. We have to show triangle formed DEF is an equilateral triangle. We know the line segment joining the mid-points of two sides of a triangle is half of the third side.

Therefore DE=

2

1

AB,EF=

2

1

BC and FD=

2

1

AC

Now, ΔABC is an equilateral triangle

⇒AB=BC=CA

⇒

2

1

AB=

2

1

BC=

2

1

CA

⇒DE=EF=FD

∴ΔDEF is an equilateral triangle.