Question

Prove that the line drawn from the mid-point of one side of a triangle parallel to
the other side, bisects the third side.

Answer 1

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

Answer:

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