Question

CM and RN are respectively the medians of $\Delta ABC$ and $\Delta PQR$. If $\Delta ABC\sim \Delta PQR$, prove that:

(A) $\Delta AMC\sim \Delta PNR$

(B) $\frac{CM}{RN}=\frac{AB}{PQ}$

Answer 1

$\left(A\right)$ Given, $△ABC$ and $△PQR$
CM is the median of $△ABC$
and RN is the median of $△PQR$

Also, $△ABC\sim △PQR$

To prove:$△AMC\sim △PNR$

Proof:
CM is the median of $△ABC$
So, $AM=MB=\frac{1}{2}AB.....\left(1\right)$
Similarly, RN is the median of $△PQR$
So, $PN=QN=\frac{1}{2}PQ.....\left(2\right)$

Given, $△ABC\sim △PQR$
$\therefore \frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}$ (Corresponding sides of similar triangle are proportional)

$\Rightarrow \frac{AB}{PQ}=\frac{CA}{RP}\Rightarrow \frac{2AM}{2PN}=\frac{CA}{RP}$ [From $\left(1\right)$ and $\left(2\right)$]

$\Rightarrow \frac{AM}{PN}=\frac{CA}{RP}.....\left(3\right)$

Also, since $△ABC\sim △PQR$

$\angle A=\angle P$ (corresponding angles of similar triangles are equal) $......\left(4\right)$

In $△AMC$ & $△PNR$,

$\angle A=\angle P$ [From $\left(4\right)$]

Also, $\frac{AM}{PN}=\frac{CA}{RP}$ [From $\left(3\right)$]

Hence, by SAS similarity,

$△AMC\sim △PNR$ $\left[Henceproved\right]$

$\left(B\right)$ $△AMC\sim △PNR$

So, $\frac{CM}{RN}=\frac{AC}{PR}=\frac{AM}{PN}$ (corresponding sides of similar triangle are proportional)

Therefore,
$\frac{CM}{RN}=\frac{AM}{PN}$

$\frac{CM}{RN}=\frac{2AM}{2PN}$

$\Rightarrow \frac{CM}{RN}=\frac{AB}{PQ}$ $\left[Henceproved\right]$