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Maths / Quadrilaterals / Converse Mid Point Theorem

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Converse Mid Point Theorem

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

GIVEN $large Delta ABC$ in which D is the mid-point of AB and $large DEparallel BC.$

To Prove E is the mid-point of AC.

Construction : Assume F to be mid point of AC. Join DF

PROOF As we have assumed F to be the mid-point of AC.

Now, in $large Delta ABC,$ D is the mid-point of AB [Given]

and F is the mid-point of AC. Therefore, by Mid Point Theorem, we have

$large DFparallel BC$ .....(i)

But we are given that , $large DEparallel BC$ ......(ii)

From (i) and (ii) , we find that two intersecting lines DE and DF are both parallel to line BC.

This is a contradiction to the parallel line axiom.

So, our supposition is wrong. Hence, E is the mid-point of AC.

Illustration:In the figure l,m and n are parallel lines intercepted by two transversal p and q such that l,m and n cut-off equal intercepts AB and BC on p. Show that l,m and n cut off equal intercepts DE and EF on q.

Solution: Join AF meetin BE at G

In $Delta$ ACF

B is the mid point of AC and BG || CF because it is given that m and n are parallel.

By Converse of Mid Point Theorem

G is the mid point of AF

In $Delta$ AFD

G is the mid point of AF and GE || AD because it is given that l and m are parallel.

By Converse of Mid Point Theorem

E is the mid point of DF

Hence DE = EF

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Converse Mid Point Theorem

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