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Mid-point theorem: - It says a line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the length of the third side.

Draw a $\vartriangle ABC$ where E and F are the mid-points of side AB and AC. Through point â€˜Câ€™ draw a line segment parallel to AB and extend EF to meet this line at point â€˜Dâ€˜

Since $AB\parallel CD$ (by construction) and ED is a transversal line then from the property of parallel lines we can say that

$\angle AEF = \angle CDF$ (Alternate angles) ...........................equation (1)

In $\vartriangle AEF$ and $\vartriangle CDF$

$\angle AEF = \angle CDF$ (From equation (1))

$\angle AFE = \angle CFD$ (Vertically opposite angle)

$AF = FC$ (As F is the midpoint of AC)

[By AAS (angle angle side) congruence rule]

So, EA=CD (By CPCT)

But EB=EA (Because E is the mid-point of AB)

Therefore EB=CD

Now in EBCD, $EB\parallel DC\& EB = DC$ (Proved above)

Thus one pair of opposite sides is equal and parallel. Hence EBCD is a parallelogram.

Since opposite sides of parallelograms are parallel.

So, $ED\parallel BC$

$\therefore EF\parallel BC$ (As F is a point on line ED)

Thus a mid-point theorem which states that a line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the length of the third side is proved.

Now we will solve another part of the question: -

In the given triangle $NO\parallel LM$, KN=2.7cm, KL=5.4cm, KO=3.9cm and we have to find OM.

As $NO\parallel LM$ we can say through mid-point theorem that N and O are the mid-points of side KL and KM

Therefore KN=NL and KO=OM

So, NL=2.7cm and OM=3.9cm

Hence final answer is OM=3.9cm

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