Question

In the given figure, $CM$ and $RN$ are respectively the medians of $△ABC$ and $△PQR$. If $△ABC\sim △PQR$, prove that:
(i) $△AMC\sim △PQR$
(ii) $CM/RN=AB/PQ$
(iii) $△CMB\sim △RNQ$.

Answer 1

Given : $△ABC\sim △PQR$ 7 $CM$ and $RN$ are medians of $△ABC$ & $△PQR$ respectively.

To prove : $\left(i\right)△AMC\sim △PNR,\left(ii\right)\frac{CM}{RN}=\frac{AB}{PQ},\left(iii\right)△CMB\sim △RNQ$

Proof : $\left(i\right)$ since $△ABC\sim △PQR.$

So, $\angle A=\angle P$

$\angle B=\angle Q,\angle C=\angle R$

& $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP},\frac{AB}{PQ}=\frac{x+x}{y+y}=\frac{2x}{2y}=\frac{x}{y}$

Let $Am=MB=x$ (Given)

$PN=NQ=y$ (Given)

In $△AMC$ & $△PNR$

$\frac{x}{y}=\frac{AM}{PN},\frac{x}{y}=\frac{AB}{PQ}=\frac{CA}{RP}$

So, $\frac{AM}{PN}=\frac{CA}{RP}$ & $\angle A=\angle P$

Therefore, $△AMC\sim △PNR$ (by side angle side similarity)

$\left(ii\right)$ As $△AMC\sim △PNR$

So, $\frac{CM}{RN}=\frac{AM}{PN}$ Proved (by similar parts of similar triangles are similar)

$\frac{CM}{RN}=\frac{AM}{PN}=\frac{x}{y}=\frac{AB}{PQ}$ Proved

$\left(iii\right)$ In $△CMB$ & $△RNQ$

$\frac{BC}{QR}=\frac{AB}{PQ}=\frac{x}{y}=\frac{BM}{QN}=\frac{x}{y}$

So, $\frac{BC}{QR}=\frac{BM}{QN},\angle B=\angle Q$

Therefore , $△CMB\sim △RNQ$

(by $SAS$ similarly )