Question

In fig., $CM$ and $RN$ are respectively then medians of $△ABC$ and $△PQR$, prove that:
$△AMC\sim △PNR$

Answer 1

$\Delta ABC$ and $\Delta PQR$

$CM$ is the median of $\Delta ABC$ and $RN$ is the median of $$\Delta PQR$

Also,

$\Delta ABC\sim \Delta PQR$

To Prove: $\Delta AMC\sim \Delta PNR$

Proof:

$CM$ is median of $\Delta ABC$

so, $AN=MB=\frac{1}{2}AB......\left(1\right)$

Similarly, $RN$ is the median of $\Delta PQR$

So, $PN=QN=\frac{1}{2}PQ......\left(2\right)$

Given,

$\Delta ABC\sim \Delta PQR$

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}$ (Corresponding sides of similar triangle are proportional)

$\frac{AB}{PQ}=\frac{CA}{RP}$

$\frac{2AM}{2PN}=\frac{CA}{RP}$ {from (1) & (2)}

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

Also,

Since $\Delta ABC\sim \Delta PQR$

$\angle A=\angle B$ (corresponding angles of similar triangles are equal)

In $\Delta AMC\sim \Delta PNR$

$\angle A=\angle P$ From (4)

$\frac{AM}{PN}=\frac{CA}{RP}$ from (3)

Hence by $S.A.S$ similarly

$\Delta AMC\sim \Delta PNR$

Hence proved.