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Explain the converse of midpoint theorem.

converse of mid-point theorem: it states that in a triangle line drawn from the mid-point of the one side of triangle, parallel to the other side intersect the third side at its mid-point.

given: ABC is a triangle. and D is the mid-point of AB.

from D a line DE is drawn parallel to BC, intersect AC at E.

TPT: E is the mid-point of AC.

construction: extend DE. from C draw a line CF parallel to BA, which intersect produced DE at F. 

proof:

since BD is parallel to CF (by the construction)

and DF is parallel to BC (given)

BDFC is a parallelogram.

BD= CF [opposite sides of the parallelogram are equal]

AD= BD [D is the mid-point of AB]

AD= CF......(1)

in the triangle AED and CEF,

∠AED=∠CEF

∠ADE=∠EFC

AD=CF  [from (1)]

therefore by the AAS congruency triangles are congruent.

thus AE = EC

i.e. E is the mid-point of AC.

hope this helps you.

cheers!!

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 The mid point theorem states that in a triange ABC , where, E is the mid point of AB and F is the mid point of  AC , EF is parallel to BC and EF is equal to half BC . The converse is that in a triangle ABC, where E is the mid point of AB , and EF is parallel to BC , then, F is the mid point of AC .
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  • 27

 The mid point theorem states that in a triange ABC , where, E is the mid point of AB and F is the mid point of  AC , EF is parallel to BC and EF is equal to half BC . The converse is that in a triangle ABC, where E is the mid point of AB , and EF is parallel to BC , then, F is the mid point of AC .

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 i need proof

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