Question

Prove that a line drawn through the mid-point
of one side of a triangle parallel to another
side bisects the third side.​

Answer 1

Answer:

proved below

Step-by-step explanation:

Given : In △ABC ,D is the mid point of AB and DE is drawn parallel to BC

To prove AE=EC

Draw CF parallel to BA to meet DE produced to F

DE∣∣BC (given)

CF∣∣BA (by construction)

Now BCFD is a parallellogram

∴BD=CF

BD=AD (as D is the mid point of AB)

⇒AD=CF

In △ADE$ and △CFE

AD=CF

∠ADE=∠CFE (alternate angles)

∠ADE=∠CEF (vertically opposite angle)

∴△ADE≅△CFE (by AAS citerion)

⇒AE=EC (by CPCT)

So E is the mid point of AC

Hence proved.

Answer 2

Given,In triangle ABC, D is the midpoint of AB such that AD=DB.

A line parallel to BC intersects AC at E as shown in above figure such that DE||BC.

To prove, E is the midpoint of AC.

Since, D is the midpoint of AB

So,AD=DB

⇒ AD/DB=1.....................(i)

In triangle ABC,DE||BC,

By using basic proportionality theorem,

Therefore, AD/DB=AE/EC

From equation 1,we can write,

⇒ 1=AE/EC

So,AE=EC

Hence, proved,E is the midpoint of AC.