## State and Prove Converse of Midpoint Theorem Class 9

**Converse of Mid Point Theorem Statement:** The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

**Given:** A ABC where E is mid point of AB, F is some point on AC & EF || BC.

**To Prove:** F is a mid point of AC.

**Construction**: Draw CM || AB, extend EF and let it cut CM at D.

**Proof:** In quadrilateral EBCD

As we know that EF || BC,

Therefore, ED || BC …….(i)

and, EB || CD (By construction) …….(ii)

Since both pairs of opposite sides are parallel. so, from eq (i) and (ii)

⇒ EBCD is a Parallelogram

Since opposite sides of a parallelogram are equal.

⇒ EB = DC

But, EB = EA (E is mid point of AB)

Hence

⇒ EA = DC ……(1)

Also,

⇒ EB||DC ( by construction AB || CD)

As ED is transversal

⇒ ∠AEF= ∠CDF (Alternate Interior Angles) …….(2)

In △AEF and △CDF

⇒ ∠AFE= ∠CFL (Vertically opposite angles)

⇒ ∠AEF= ∠CDF (From (2))

⇒ AE = CD (From (1))

△AEF ≅ △CDF (AAS rule)

So, AF = CF (C.P.C.T.)

Hence, F is a mid point of AC

Hence proved.

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