Question

Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (Using basic proportionality theorem).

Answer 1

 Given that ABC is a traingle and D and E are mid points of traingle. Construction : Join C and D , B and E. Prove that : DE // BC.Proof :  In ΔADE = ΔBDE AD = BD ( D is mid point) DE = DE ( same height) Area of ΔADE = Area of ΔBDE -------------(1) In ΔADE = ΔCDE AE = EC ( E is mid point) DE = DE ( same height) Area of ΔADE = Area of ΔCDE -------------(2)from (1) and (2) we get  Area of ΔBDE = Area of ΔCDE ∴ DE // BC.

Answer 2

$\bf\huge\underline\mathtt\red{Question}$

Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (Using basic proportionality theorem).

$\bf\huge\underline\mathtt\blue{Answer}$

We have ∆ABC, in which D is the midpoint of AB and E is a point on AC such that DE || BC.

Since, DE || BC [given]

Therefore, Using the Basic proportionality theorem,

We get

$\dfrac{AD}{DB}$ = $\dfrac{AE}{EC}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀.....(1)

But D is the mid-point of AB

Therefore, AD = DB

=> $\dfrac{AD}{DB}$ = 1 ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ .....(2)

From (1) and (2),

1 = $\dfrac{AE}{EC}$ => EC = AE

=> E is the mid point of AC. Hence, it is proved that a line through the midpoint of one side of a triangle parallel to another side bisects the third side.