Theorem 8.9 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side Class 9

Theorem 8.9 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Given: ABCD is a triangle where E and F are midpoints of AB and AC respectively.

To prove: EF||BC

Construction: draw a line segment through C parallel to AB and extend EF to meet this line at D.

Proof: since CD || AB with transversal ED.    (By construction )

In △EAF and △FCD

∠EAF = ∠FCD  (Alternate Interior Angle)

  AF =FC          (F is the midpoint of AC)

∠AEF = ∠FDC   (Alternate Interior Angle)

△AEF △FCD (By Using ASA Property)

So, AE = CD (By CPCT)

But, AE = EB (E is the midpoint of AB)

Then, EB = DC

If in a quadrilateral one pair of opposite sides are equal and parallel then it is a parallelogram.

Hence, BCDE is a parallelogram.

Since opposite sides of a parallelogram are parallel

So, ED || BC

It means, EF || BC 

Hence, proved.

1 thought on “Theorem 8.9 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side Class 9”

Leave a Reply

%d bloggers like this: