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.
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