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

Given: AM and DN are the medians of triangles ABC and DEF respectively and $$\triangle$$ ABC ~ $$\triangle$$ DEF.

We have to prove: Area($$\triangle$$ ABC)/Area($$\triangle$$ DEF) = AM2/DN2

Since, $$\triangle$$ ABC ~ $$\triangle$$ DEF (Given)
Area($$\triangle$$ ABC)/Area($$\triangle$$ DEF) = $$(AB^2/DE^2)$$ ……………………………(i)
and, AB/DE = BC/EF = CA/FD ………………………………………(ii)
$${{AB} \over {DE}} ={{{{1} \over {2}}BC} \over {{{1} \over {2}}EF}} = {{CD} \over {FD}}$$

In $$\triangle$$ ABM and $$\triangle$$ DEN,

Since $$\triangle$$ ABC ~ $$\triangle$$ DEF
$$\angle$$ B = $$\angle$$ E
AB/DE = BM/EN [Already Proved in equation (i)]
$$\triangle$$ ABC ~ $$\triangle$$ DEF [SAS similarity criterion]
AB/DE = AM/DN …………………………………………………..(iii)
$$\triangle$$ ABM ~ $$\triangle$$ DEN

As the areas of two similar triangles are proportional to the squares of the corresponding sides.
area($$\triangle$$ ABC)/area($$\triangle$$ DEF) = $$AB^2/DE^2 = AM^2/DN^2$$
Hence, proved.