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

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To do:

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

Solution:

Let $\triangle ABC \sim \triangle DEF$.

$AP$ and $DQ$ are medians.

We know that,

The areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

This implies,

$\frac{\operatorname{ar} \Delta \mathrm{ABC}}{\operatorname{ar} \Delta \mathrm{DEF}}=\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}$

$\Delta \mathrm{ABC} \sim \Delta \mathrm{DEF}$

This implies,

$\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}$

$=\frac{2 \mathrm{BP}}{2 \mathrm{EQ}}$

$=\frac{\mathrm{BP}}{\mathrm{EQ}}$

$\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BP}}{\mathrm{EQ}}$..........(i)

$\angle \mathrm{B}=\angle \mathrm{E}$            (Corresponding angles)

Therefore, by SAS criterion,

$\Delta \mathrm{ABP} \sim \Delta \mathrm{DEQ}$

This implies,

$\frac{\mathrm{BP}}{\mathrm{EQ}}=\frac{\mathrm{AP}}{\mathrm{DQ}}$........(ii)

From (i) and (ii), we get,

$\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AP}}{\mathrm{DQ}}$

Therefore,

$\frac{\text { ar } \triangle \mathrm{ABC}}{\operatorname{ar} \Delta \mathrm{DEF}}=\frac{\mathrm{AP}^{2}}{\mathrm{DQ}^{2}}$

Hence proved.

Updated on 10-Oct-2022 13:21:22