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Step-by-step explanation:
According to this theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Proof:
Suppose a line DE, intersects the two sides of a triangle AB and AC at D and E, such that;
AD/DB = AE/EC ……(1)
Assume DE is not parallel to BC. Now, draw a line DE’ parallel to BC.
Hence, by similar triangles,
AD/DB = AE’/E’C ……(2)
From eq. 1 and 2, we get;
AE/EC = AE’/E’C
Adding 1 on both the sides;
AE/EC + 1 = AE’/E’C +1
(AE +EC)/EC = (AE’+E’C)/E’C
AC/EC = AC/E’C
So, EC = E’C
This is possible only when E and E’ coincides.
But, DE’//BC
Therefore, DE//BC.
Hence, proved.
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