Question

state and prove mid point theorem

Answer 1

Statement – “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 of the Mid- Point Theorem

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

DE is parallel to BC

DE ∥∥ BC

DE = (1/2 *  BC).

Now consider the below figure,

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 Proved.

The Mid- Point Theorem can also be proved by the use of triangles. The line segment which is on the angle, suppose two lines are drawn in parallel to the x and the y-axis which begin at endpoints and also the midpoint, then the result is said to be two similar triangles. This relation of these triangles forms the Mid- Point Theorem





















by byjus

Answer 2

Mid point theorem states that a line drawn through the midpoint of one side of a triangle parallel to the second side bisects the third side

Take a triangle ABC. Locate D as midpoint of AB and extend a line parallel to side BC to intersect side CA at E
Now, DE||BC
So, $\frac{AD}{DB} = \frac{AE}{EC}$ (Basic Proportionality Theorem)
      But, AD=DB(Midpoint)
So, $\frac{AE}{EC} =1$
  SO we get that AE=EC