Question

state and prove mid point theorem class 9th​

Answer 1

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.”

Step-by-step explanation:

Proof:

If the line segment adjoins midpoints of any of the sides of a triangle, then the line segment is said to be parallel to all the remaining sides, and it measures about half of the remaining sides.

Consider the triangle ABC,

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 II BC

DE = (1/2 * BC).

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)

EF = DE (by construction)

By SAS congruence criterion,

Δ CFE ≅ Δ 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 that 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 that 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).

Therefore, the Midpoint theorem is proved.

Answer 2

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.

Take a triangle ABC,E and F are the mid-points of side AB and AC resp.

Construction:-Through C,draw a line II BA to meet EF produced at D.

Proof:-

In Triangle AEF and CDF

1.AF=CF(F is midpoint of AC)

2.<AFE=<CFD (Vertically opp. angles)

3.<EAF=<DCF [Alt. angles,BA II CD(by construction) and AC is a transversal]

4.So,Triangle AEF = CDF(ASA)

5.EF=FD AND AE = CD (c.p.c.t)

6.AE=BE(E is midpoint of AB)

7.BE=CD(from 5 and 6)

8.EBCD is a IIgm [BA II CD (by construction) and BE = CD(from 7)]

9.EF II BC AND ED=BC (Since EBCD is a IIgm)

10.EF = 1/2 ED (Since EF = FD,from 5)

11.EF = 1/2 BC (Since ED = BC,from 9)

Hence,EF II BC AND EF = 1/2 BC which proves the mid-point theorem.