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It is given that sides $AB$ and $AC$ and median $AD$ of $\Delta ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another $\Delta PQR$. Show that $\Delta ABC \sim \Delta PQR$.
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Extend $AD$ to $E$ and $PM$ to $N$ such that $AD = DE$ and $PM = MN$.
Proving $\Delta DAB \cong \Delta DEC$ (By SAS congruency criterion)
Similarly, $\Delta MPQ \cong \Delta MNR$
$\therefore AB = CE$ and $PQ = NR$ (by cpct)
Given $\frac{AB}{PQ} = \frac{AD}{PM} = \frac{AC}{PR} \Rightarrow \frac{CE}{NR} = \frac{AE/2}{PN/2} = \frac{AC}{PR} \Rightarrow \frac{CE}{NR} = \frac{AE}{PN} = \frac{AC}{PR}$
Hence $\Delta CAE \sim \Delta RPN$ (By SSS similarity criterion)
$\Rightarrow \angle 1 = \angle 2$, similarly $\angle 3 = \angle 4$
Adding, we get $\angle 1 + \angle 3 = \angle 2 + \angle 4$ or $\angle BAC = \angle QPR$
Hence, $\Delta ABC \sim \Delta PQR$ (By SAS similarity criterion)
Proving $\Delta DAB \cong \Delta DEC$ (By SAS congruency criterion)
Similarly, $\Delta MPQ \cong \Delta MNR$
$\therefore AB = CE$ and $PQ = NR$ (by cpct)
Given $\frac{AB}{PQ} = \frac{AD}{PM} = \frac{AC}{PR} \Rightarrow \frac{CE}{NR} = \frac{AE/2}{PN/2} = \frac{AC}{PR} \Rightarrow \frac{CE}{NR} = \frac{AE}{PN} = \frac{AC}{PR}$
Hence $\Delta CAE \sim \Delta RPN$ (By SSS similarity criterion)
$\Rightarrow \angle 1 = \angle 2$, similarly $\angle 3 = \angle 4$
Adding, we get $\angle 1 + \angle 3 = \angle 2 + \angle 4$ or $\angle BAC = \angle QPR$
Hence, $\Delta ABC \sim \Delta PQR$ (By SAS similarity criterion)