Question

The mid-point theorem and prove it

Answer 1

Answer:

The mid point of two sides of triangle then the line segment thus formed in parallel to the third sides of triangle and also the line segment of the half of the length of the third sides .This theorem is called mid point theorem .....

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Answer 2

MidPoint Theorem Statement.....

The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”

MidPoint Theorem Proof........

If midpoints of any of the sides of a triangle are adjoined by the line segment, then the line segment is said to be in parallel to all the remaining sides and also will measure about half of the remaining sides.

Consider the triangle ABC, as shown in the above figure,

Let E and D be the midpoints of the sides AC and AB. Then the line DE is said to be parallel to the side BC, whereas the side DE is half of the side BC; i.e.

DE∥BC

DE = (1/2 * BC).

Now consider the below figure,

Mid- Point Theorem

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

In the triangle, ADE, and also the triangle CFE

EC= AE —– (given)

∠CEF = ∠AED {vertically opposite angles}

EF = DE { by construction}

hence,△ CFE ≅ △ ADE {by SAS}

Therefore,

∠CFE = ∠ADE {by c.p.c.t.}

∠FCE= ∠DAE {by c.p.c.t.}

and CF = AD {by c.p.c.t.}

The angles, ∠CFE and ∠ADE are the alternate interior angles. Assume CF and AB as two lines which are intersected by the transversal DF.

In a similar way, ∠FCE and ∠DAE are the alternate interior angles. Assume CF and AB are the two lines which are intersected by the transversal AC.

Therefore, CF ∥ AB

So, CF ∥ BD

and CF = BD {since BD = AD, it is proved that CF = AD}

Thus, BDFC forms a parallelogram.

By the use of properties of a parallelogram, we can write

BC ∥ DF

and BC = DF

BC ∥ DE

and DE = (1/2 * BC).

Hence, the midpoint theorem is Proved......

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