If $AD$ and $PM$ are medians of $\vartriangle ABC$ and $\vartriangle PQR$ respectively where, $\vartriangle ABC\sim \vartriangle PQR$. Prove that $\frac{AB}{PQ}=\frac{AD}{PM}$.

Given: $AD$ and $PM$ are medians of $\vartriangle ABC$ and $\vartriangle PQR$ respectively where, $\vartriangle ABC\sim \vartriangle PQR$. 
 
To do: To prove that $\frac{AB}{PQ}=\frac{AD}{PM}$.

Solution:




It is given that $\vartriangle ABC \sim   \vartriangle PQR$

As known that the corresponding sides of similar triangles are always proportional.

$\therefore  \frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}\ ...\ ( i)$

Also, $\angle A = \angle P,\ \angle B = \angle Q,\ \angle C = \angle R\ ...\ ( ii)$

$\because AD$ and $PM$ are medians, so they divide their opposite sides $BC$ and $QR$ respectively.

$\therefore  BD=\frac{BC}{2}$ and $QM=\frac{QR}{2}\ ...\ ( iii)$

From equations $( i)$ and $(  iii)$, we get

$\frac{AB}{PQ}=\frac{BD}{QM}\ ...\ ( iv)$

In $\vartriangle ABD$ and $\vartriangle PQM$,

$\angle B = \angle Q$         [Using equation $( ii)$]

$\frac{AB}{PQ}=\frac{BD}{QM}$     [Using equation $( iv)$]

$\therefore  \vartriangle ABD \sim   \vartriangle PQM$  [By SAS similarity criterion)]

$\Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$
 

Updated on: 10-Oct-2022

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