ong>Answer:

the mid point theorem states that the line segment in a triangle joining the midpoints of two sides of the triangle is said to be parallel to its third side and is also half of third side

Now prove :

refer to the diagram...

let E and D be the midpoints of sides AC and AB

then the line DE is said to be parallel to the third side BC whereas the side DE is half of the side BC I. e

DE parallel to BC

DE = 1 / 2 * BC

Construction = Extend the line segment DE and produce it to F such that EF = DE

In triangle ADE and CFE

EC = AE ( given)

angle CEF = angle AED ( given)

EF = DE ( construction)

By SAS CONGRUENCE criterion

angleCFE congruence to angle ADE

therefore ,

angleCFE = angleADE ( by cpct)

angleFCE = angleDAE ( by cpct)

and CF = AD ( by cpct)

angleCFE and angleADE are alternative interior angles

Assume CF and AB as two lines which are interesected by the transversal DE

In a similar way angleFCE and angleDAE are alternative interior angles

Assume CF and AB are the two lines which are interesected by the transversal AC

therefore,

CF parallel to AB

so , CF parallel to BD

And CF = BD ( since BD = AD, it is proved that CF = AD)

Thus, BDFC forms a parallelogram.

By PROPERTIES of parallelogram

we can write,

BC parallel to DF

And BC = DF

BC parallel to DE

And DE = 1/2BC

Hence midpoint theorem proved .

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