Question

explain mid point theorem with proof​

Answer 1

Answer:

Mid - Point Theorem

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 triangle.”

Proof:

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 sides BC, whereas the side DE is half of the side BC; i.e.

DE is parallel to BC

DE = (1/2*BC).

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)

angle CEF = angle AED {vertically opposite angles}

EF = DE { by construction}

hence,

triangle CFE is congruent to triangle ADE {by SAS}

Therefore,

angle CFE = angle ADE {by c.p.c.t.}

angle FCE = angle DAE {by c.p.c.t.}

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

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

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

Therefore, CF is parallel to AB

So, CF is parallel to 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 is parallel to DF

and BC = DF

BC is parallel to DE

and DE = (1/2 * BC).

Hence, the midpoint theorem is Proved.

Answer 2

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.