Question

using theorem 6.1 prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side​

Answer 1

Proved below.

Step-by-step explanation:

Given:  

As shown in the figure, let in Δ ABC, D is midpoint of AB and DE is parallel to BC.

∴ AD = DB

To prove:  

AE = EC

Proof:  

Since,D║BC

∴ By Basic Proportionality Theorem,

$\frac{AD}{DB}=\frac{AE}{EC}$

Since, AD = DB          [given]

∴ $\frac{AE}{EC}=1$  

∴ AE = EC

Hence proved.

Answer 2

Answer:

Step-by-step explanation:

REFER to the photo given below and don't FORGET to mark me as BRAINLIEST.

HOPE you understand