Question

$CM$ and $RN$ are respectively, the medians of $\Delta ABC$ and $\Delta PQR$. If $\Delta ABC\sim \Delta PQR,$
How many of the following satatements are true:
i) $\Delta AMC\sim \Delta PNR$
ii) $\frac{CM}{RN}=\frac{AB}{PQ}$
iii) $\Delta CMB\sim \Delta RNQ$

Answer 1

$\left(i\right)\because △ABC\sim △PQR$

$⟹\frac{AB}{PQ}=\frac{BC}{RQ}=\frac{AC}{PR}-\left(a\right)$

$⟹\frac{2AM}{2PN}=\frac{AC}{PR}$

$⟹\frac{AM}{PN}=\frac{AC}{PR}-\left(1\right)$

$\because trianglesweresimilar$

$⟹\angle BAC=\angle QPR$

$\therefore In△AMC\&△PNR$

$⟹\angle MAC=\angle BAC=\angle NPR=\angle QPR$

$⟹\angle MAC=\angle NPRand\frac{AM}{PN}=\frac{AC}{PR}-\left(2\right)$

$\therefore △AMC\sim △PNR$

$⟹\frac{AM}{PN}=\frac{AC}{PR}=\frac{CM}{RN}-\left(3\right)$

$\left(ii\right)\because trianglesaresimilar$

$impliesfrom\left(2\right)$

$\frac{AB}{PQ}=\frac{AC}{PR}$

$from\left(2\right)$

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AM}{PN}$

$from\left(3\right)$

$\frac{AB}{PQ}=\frac{CM}{RN}$

$\left(iii\right)In△CMB\&△RNQ$

$\frac{CM}{RN}=\frac{AB}{PQ}$

$from\left(a\right)$

$\frac{CM}{RN}=\frac{AB}{PQ}=\frac{BC}{RQ}$

$⟹\frac{CM}{RN}=\frac{BC}{RQ}=\frac{2MB}{2NQ}$

$⟹\frac{CM}{RN}=\frac{BC}{RQ}=\frac{MB}{NQ}$

$\therefore △CMB\sim △RNQ$

$\therefore Allthreestatementsaretrue.$