Question

In the given figure, seg PD is a median of $∆$ PQR. Point T is the mid point of seg PD. Produced QT intersects PR at M. Show that $\frac{\mathrm{PM}}{\mathrm{PR}}$= $\frac{1}{3}$.
[Hint: DN || QM]

Answer 1

PD is the median of QR.
So, D is the midpoint of QR.
DN is drawn parallel to QM.
By converse of midpoint theorem, N is the midpoint of MR. .....(1)
Similarly, T is the midpoint of PD
Also, DN || QM
So, By converse of midpoint theorem,
M is the midpoint of PN. .....(2)
From (1) and (2) we have
PM = MN = NR
$$\begin{gathered} \Rightarrow \frac{\mathrm{PM}}{\mathrm{PR}}=\frac{\mathrm{PM}}{\mathrm{PM}+\mathrm{MN}+\mathrm{NR}}=\frac{\mathrm{PM}}{\mathrm{PM}+\mathrm{PM}+\mathrm{PM}}=\frac{1}{3\mathrm{PM}}=\frac{1}{3} \\ \Rightarrow \frac{\mathrm{PM}}{\mathrm{PR}}=\frac{1}{3} \end{gathered}$$
Hence Proved.