**Question:**

Prove that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles.

**Solution:**

∆*ABC* is shown below. *D, E* and *F* are the midpoints of sides *AB, BC* and *CA*, respectively.

As,* D* and * E* are the mid points of sides *AB*, and *BC* of ∆ *ABC.*

∴ *DE ∣∣ AC* (By midpoint theorem)

Similarly, *DF ∣∣ BC* and *EF ∣∣ AB*.

Therefore, *ADEF, BDFE* and *DFCE *are all parallelograms.

Now, *DE* is the diagonal of the parallelogram *BDFE*.

∴ ∆*BDE ≅ ∆FED*

Simiilarly, *DF* is the diagonal of the parallelogram *ADEF*.

∴ ∆

And,

*DAF**≅ ∆FED*And,

*EF*is the diagonal of the parallelogram*DFCE*.∴ ∆

So, all the four triangles are congruent.

*EFC ≅ ∆FED*So, all the four triangles are congruent.

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