Question

Prove that ratio of areas of two similar triangles is equal to the square of ratio of their corresponding medians.

Answer 1

Similar triangles: Those triangles whose size can be changed but shape is same is known as similar triangles.
Median: Line segment from vertex to the opposite which bisects that side is median. Every triangle has 3 medians.

Complete step-by-step answer:
Let us take two triangles $ABC$ and $PQR$ where $AD$ is the median in triangle $ABC$ and
$PM$ is the median in the triangle $PQR$. We have to prove that

$\dfrac{{AreaOf\Delta ABC}}{{AreaOf\Delta PQR}} = {\left( {\dfrac{{AD}}{{PM}}} \right)^2}$
Proof: Since we are given that two triangles $ABC$ & $PQR$ are similar
$\therefore$ $\Delta ABC \sim \Delta PQR$ (given)
$\Rightarrow$ $\angle B = \angle Q$ (Corresponding angles of similar triangles are also equal)
$\Rightarrow$ $\dfrac{{AB}}{{PQ}} = \dfrac{{BC}}{{QR}}$ (Corresponding scales of similar triangles of same proportion)
\Rightarrow $$$\dfrac{{AB}}{{PQ}} = \dfrac{{BD + DC}}{{QM + MR}} = \dfrac{{BD + BD}}{{QM + QM}}$$ \because $$AD&PM$are medians So$BD = DC&QM = MR$$ \Rightarrow $$$\dfrac{{AB}}{{PQ}} = \dfrac{{2BD}}{{2QM}} = \dfrac{{BD}}{{QM}}-------(1) $\therefore $ In $\Delta ABD$ & $\Delta PQM$ $\angle B = \angle Q$ (already proved) $\dfrac{{AB}}{{PQ}} = \dfrac{{BD}}{{QM}}$ (From (1)) ----------(2) $\therefore $ $\Delta ABD \sim \Delta PQM$ $\dfrac{{AB}}{{PQ}} = \dfrac{{DM}}{{PM}}$ Corresponding sides of similar $\Delta $’s in same proportion \dfrac{{AreaOf\Delta ABC}}{{AreaOf\Delta PQR}} = {\left( {\dfrac{{AB}}{{PQ}}} \right)^2}$$\dfrac{{AreaOf\Delta ABC}}{{AreaOf\Delta PQR}} = {\left( {\dfrac{{BD}}{{QM}}} \right)^2}$$-----(from equation (2))

Note: 1) From this we can also prove that the ratio of areas of $2$ similar $\Delta$’s is equal to the square of ratio of their altitude.
2) Similarity of triangles can be proved by two methods by taking angles and sides.