**Theorem 8.9 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.**

**Given:** ABCD is a triangle where E and F are midpoints of AB and AC respectively.

**To prove:** EF||BC

**Construction:** draw a line segment through C parallel to AB and extend EF to meet this line at D.

**Proof:** since CD || AB with transversal ED. (By construction )

In △EAF and △FCD

∠EAF = ∠FCD (Alternate Interior Angle)

AF =FC (F is the midpoint of AC)

∠AEF = ∠FDC (Alternate Interior Angle)

**∴** △AEF **⩭** △FCD (By Using ASA Property)

So, AE = CD (By CPCT)

But, AE = EB (E is the midpoint of AB)

Then, EB = DC

If in a quadrilateral one pair of opposite sides are equal and parallel then it is a parallelogram.

Hence, BCDE is a parallelogram.

Since opposite sides of a parallelogram are parallel

So, ED || BC

It means, EF || BC

Hence, proved.

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