prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that your have proved it in class IX)

To prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side, we can use the concept of similar triangles.

Let's consider a triangle ABC, where D is the mid-point of side AB and EF is drawn parallel to side BC, intersecting side AC at point E and side AB at point F.

To prove that EF bisects side AC, we need to show that AE = EC.

First, we can establish that triangle AEF is similar to triangle ABC. This is because angle AEF is equal to angle ABC (as they are corresponding angles formed by parallel lines), and angle EAF is equal to angle BAC (as they are alternate interior angles formed by parallel lines).

Since triangle AEF is similar to triangle ABC, we can use the property of similar triangles that states the ratio of corresponding sides is equal. In this case, we can write:

AE/AB = EF/BC

Since D is the mid-point of side AB, we know that AD = DB. Therefore, AB = 2AD.

Substituting this into the equation, we get:

AE/2AD = EF/BC

Cross-multiplying, we have:

AE * BC = 2AD * EF

Since EF is parallel to BC, we can use the alternate interior angles formed by parallel lines to establish that angle EFC is equal to angle ABC.

Using the property of corresponding angles in similar triangles, we can write:

EF/BC = FC/AC

Substituting this into the equation, we get:

AE * BC = 2AD * FC/AC

Cross-multiplying again, we have:

AE * AC = 2AD * FC

Since AD = DB, we can substitute this into the equation:

AE * AC = 2DB * FC

But we know that DB = AD, so we can simplify further:

AE * AC = 2AD * FC = AB * FC

Since AE * AC = AB * FC, we can conclude that AE = EC.

Therefore, the line EF, drawn through the mid-point of side AB parallel to side BC, bisects side AC.