Given \(\Delta ABC\) is similar to \(\Delta PQR\)., and medians
\(AD\) and
\(PM\) drawn to the corresponding sides. Prove that the medians are proportional to the sides from which they are drawn i.e., \(\frac{AB}{PQ} = \frac{AD}{PM}\).
Proof:
Since \(AD\) is the median of \(\Delta ABC\)
\(BD =\) \( = \frac{1}{2}\)
Similarly, \(PM\) is the median \(\Delta PQR\)
\(QM = \) \(= \frac{1}{2}\)
Given \(\Delta ABC \sim \Delta PQR\).
Corresponding sides of similar triangle are proportional.
So, \(\frac{AB}{PQ} = \frac{BC}{QR}\)
\(\frac{AB}{PQ} = \frac{2BD}{2QM}\) (since \(AD\) and \(PM\) are medians)
\(\frac{AB}{PQ} = \frac{BD}{QM}\) - - - - - (1)
Since \(\Delta ABC \sim \Delta PQR\)
Corresponding angles of similar triangles are equal.
\(\angle B = \angle\) - - - - - (2)
Now, in \(\Delta ABD\) and \(\Delta PQM\)
\(\angle B = \angle Q\) (from (2))
\(\frac{AB}{PQ} = \frac{BD}{QM}\) (from (1))
Thus, \(\Delta ABD \sim \Delta PQM\) (by )
Since corresponding sides of similar triangles are proportional.
\(\frac{AB}{PQ} = \frac{AD}{PM}\)
Hence proved.