Theorem 8.10 Class 9 Maths Explanation with Proof

Today, we will discuss theorem 8.10 class 9 Maths which is related to Chapter 8 Quadrilaterals Class 9 Mathematics. After understanding theorem 8.10, you can solve the exercise questions given in the NCERT book of Class 9 Maths.

 

 Theorem 8.10 Class 9

The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

Given

In an ∆ABC, E is the midpoint of AB and EF∥BC.

To Prove

F is the midpoint of AC

Construction 

Draw CM∥BA and extend EF such that it intersects CM at D.

Proof 

In ∆AEF and ∆CDF 

∠FAE=∠FCD(∵AB∥CM and AC is a transversal, so alternate angles  are equal) 

∠AEF=∠FDC(∵AB∥CM and ED is a transversal, so alternate angles are equal) 

CD=AE(∵BCDE is a parallelogram,so CD=BE=AE)

∆AEF=∆CDF (By ASA congruence rule)

Then AF=CF( By CPCT)

Hence, F is a mid-point of AC

The converse of the mid-point theorem can also be proved by another method as under:

Let F not be the midpoint of AC and D be the midpoint of AC. Join ED.

In ∆ABC, E is the midpoint of AB and D is the midpoint of AC, then by the mid-point theorem,

ED∥BC ………(i)

But EF∥BC ………(Given)(ii)

From equations (i) and (ii), we have that ED and EF both are parallel to BC. But this is a contradiction to the parallel line axiom. So our assumption that F is not the midpoint of AC is wrong.

Hence, F is the midpoint of AC

Hence proved

Related Topics

1. Theorem 8.1

2. Theorem 8.9