Math, asked by AnshikaTallented, 10 months ago

mid point theorm explain the mid point theorem of class 9 chapter number 7 triangles ​

Answers

Answered by Anonymous
0

Answer:

A midpoint is a point on a line segment equally distant from the two endpoints. The Midpoint Theorem is used to make a bold statement regarding triangle sides and their lengths. Given a triangle, if we connect two sides with a line segment, and this line segment joins each of the two sides at the centers, or midpoints of each side, we can know two very important aspects about the triangle and the relationships between the sides.

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

Anytime you have a line segment that connects two sides of a triangle at the midpoints, you automatically know that the sides are cut in half, and that the segment is parallel to the third side of the triangle. Parallel sides are shown by using this symbol ||. You also know the line segment is one-half the length of the third side.

Answered by nilesh102
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.”

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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