Math, asked by ak1068454, 2 months ago

prove Mid Point therom​

Answers

Answered by Anonymous
5

Step-by-step explanation:

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 triangle ADE and CFE,

EC = AE —– (given)

∠CEF = ∠AED (vertically opposite angles)

∠DAE = ∠ECF (alternate angles)

By ASA congruence criterion,

△ CFE ≅ △ ADE

Therefore,

∠CFE = ∠ADE {by c.p.c.t.}

∠FCE= ∠DAE {by c.p.c.t.}

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

∠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 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.

Answered by jahnvisharma42
1

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

In triangle ADE and CFE,

EC = AE —–   (given)

∠CEF = ∠AED (vertically opposite angles)

∠DAE = ∠ECF (alternate angles)

By ASA congruence criterion,

△ CFE ≅  △ ADE

Therefore,

∠CFE = ∠ADE {by c.p.c.t.}

∠FCE= ∠DAE    {by c.p.c.t.}

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

∠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 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.

MidPoint Theorem Formula

In Coordinate Geometry, the midpoint theorem refers to the midpoint of the line segment. It defines the coordinate points of the midpoint of the line segment and can be found by taking the average of the coordinates of the given endpoints. The midpoint formula is used to determine the midpoint between the two given points.

If P1(x1, y1) and P2(x2, y2) are the coordinates of two given endpoints, then the midpoint formula is given as:

Midpoint = [(x1 + x2)/2, (y1 + y2)/2]

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