Question

✳✳derive the Mid - point theorem
class - 10
chapter - Triangles
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⭐⭐⭐quality required ⭐⭐⭐

Answer 1

"In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it."Explanation:
If the midpoints of any sides of a triangle are joined by a line segment, then this line segment will be parallel to the remaining side and will measure half of the remaining side.

Let us consider ABC is a triangle as shown in the following figure:
Mid Point Theorem

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

DE ∥ BC

DE = 12 BC.

Midpoint theorem plays a vital role in mathematics. Let us learn more about midpoint theorem, its converse and its applications in detail.

Answer 2

Geometry is one among the fundamental and essential branches of mathematics. This field deals with the geometrical problems and figures which are based on the properties. One of the important theorems in the field of geometry that deals with the properties of triangles are called as the Mid- Point Theorem.

The theory of midpoint theorem is used in the coordinate geometry stating that the midpoint of the line segment is an average of the endpoints. To solve an equation by the use of this Theorem, it is important that both the ‘x’ and the ‘y’ coordinates should be known. The Mid- Point Theorem is also useful in the fields of calculus and algebra.

Statement and Proof of Mid- Point Theorem

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,





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Mid – Point Theorem

Mid – Point Theorem

February 14, 20181 Comment

Geometry is one among the fundamental and essential branches of mathematics. This field deals with the geometrical problems and figures which are based on the properties. One of the important theorems in the field of geometry that deals with the properties of triangles are called as the Mid- Point Theorem.

The theory of midpoint theorem is used in the coordinate geometry stating that the midpoint of the line segment is an average of the endpoints. To solve an equation by the use of this Theorem, it is important that both the ‘x’ and the ‘y’ coordinates should be known. The Mid- Point Theorem is also useful in the fields of calculus and algebra.

Statement and Proof of Mid- Point Theorem

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.