Theorem 8.9 Class 9 Maths Explanation with Proof

Today, we will discuss theorem 8.9 class 9 Maths which is related to Chapter 8 Quadrilaterals Class 9 Mathematics. After understanding theorem 8.9, you can solve the exercise questions given in the NCERT book of Class 9 Maths.

 Theorem 8.9 Class 9

The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Given 

A ∆ABC  in which D and E are the midpoints of sides AB and AC resepectively.DE is joined.

To Prove

`DE∥BC and DE=frac{1}{2} BC`

 

Construction 

Produce the line segment DE to F, such that DE=EF. Join FC.

Proof 

In ∆s AED and CEF, we have 

AE=CE(∵E is  the midpoint of AC)

∠AED=∠CEF (Vertically opposite angles)

And, DE=EF(By construction)

So, by SAS criterion of congruence, we have

∆AED≅∆CEF 

⟹AD=CF (CPCT)………..(i)

And , ∠ADE=∠CFE ……….(ii)

Now, D is the midpoint of AB

⇒AD=DB 

⇒DB=CF  [from(i)AD=CF]………(iii)

Now, DF intersects AD and FC at D and F

 respectively such that

∠ADE=∠CFE                         [from (ii)]

Alternate interior angles are equal.

∴AD∥FC 

⇒DB∥CF 

From (iii) and (iv), we find that DBCF is quadrilateral such that one pair of sides are equal and parallel.

∴DDBCF is a parallelogram 

⇒DF∥BC and DF=BC (∵opposite sides of a ∥gm are equal and parallel)

But D,E, F are collinear and DE=EF

`DE∥BC and DE=frac{1}{2} BC`

Hence proved

Related Topics

1. Theorem 8.10

2. Theorem 8.8