Converse of Mid Point Theorem 8.10 Class 9

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: In △ ABC where E is the 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.