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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.

Q6   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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Let AD and PS be medians of both similar triangles.

$\triangle ABC\sim \triangle PQR$

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}............................1$

$\angle A=\angle P,\angle B=\angle Q,\angle \angle C=\angle R..................2$

$BD=CD=\frac{1}{2}BC\, \, and\, QS=SR=\frac{1}{2}QR$

Purring these value in 1,

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AC}{PR}..........................3$

In $\triangle ABD\, and\, \triangle PQS,$

$\angle B=\angle Q$    (proved above)

$\frac{AB}{PQ}=\frac{BD}{QS}$      (proved above)

$\triangle ABD\, \sim \triangle PQS$     (SAS )

Therefore,

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AD}{PS}................4$

$\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{AB^2}{PQ^2}=\frac{BC^2}{QR^2}=\frac{AC^2}{PR^2}$

From 1 and 4, we get

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{AD}{PS}$

$\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{AD^2}{PS^2}$

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