**Q) In 𝛥ABC, D, E and F are midpoints of BC,CA and AB respectively. Prove that △ 𝐹𝐵𝐷 ∼ △ DEF and △ DEF ∼ △ ABC**

**Ans: **

**(i): Prove that △ 𝐹𝐵𝐷 ∼ △ DEF:**

Let’s start from comparing triangles △ FBD and △ DEF.

Since by midpoint theorem, The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.

Therefore, in the given diagram: FE ǁ BC

∴ ∠ BFD = ∠ EDF (Alternate interior angles)

∠ BDF = ∠ EFD (Alternate interior angles)

∴ by AA similarity criterion, we get:

**△ FBD ~ △ DEF …. Hence Proved !**

**(ii) Prove that △ DEF ∼ △ ABC:**

In the given diagram, D, E and F are mid points of BC, AC and AB respectively,

By mid point theorem, FE = , FD = and ED = ,

∴ ,

∴ △ ABC ~ △ DEF

**∴ △ DEF ~ △ ABC …. Hence Proved !**

**Please do press “Heart” button if you liked the solution. **