Establish that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Proof:
Let us assume \(\Delta ABC\), where \(DE\) is parallel to \(BC\) and \(D\) is the mid point of \(AB\).
To prove that \(E\) is the mid point of \(AC\).
In \(\Delta ABC\), \(DE || BC\)
Then by,
\(\frac{AD}{DB} = \)
Since \(D\) is the mid-point of \(AB\), \(AD = \)
\(= \frac{AE}{EC}\)
\(= \frac{AE}{EC}\)
\(EC = AE\)
Thus, \(E\) is the mid-point of \(AC\).
Hence proved.
Answer variants:
\(\frac{DB}{DB}\)
\(\frac{AE}{EC}\)
\(AE\)
\(\frac{AB}{BC}\)
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Converse of basic proportionality theorem
Basic proportionality theorem