Verify that a line passing through the midpoint of one side of a triangle and parallel to another side divides the third side into two equal parts.
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:
Converse of basic proportionality theorem
1
Basic proportionality theorem
\(AE\)
\(\frac{AB}{BC}\)
\(\frac{DB}{DB}\)
\(\frac{AE}{EC}\)