Sides $A B, A C$ and median $A D$ of a triangle $A B C$ are respectively proportional to sides $P Q, P R$ and median $P M$ of another triangle $P Q R$ . Show that $△ A B C \sim △ P Q R$ .

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Solution

$g i v e n : \frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M} = k (l e t)$

$A B = k P Q - (i)$

$A C = k P R - (i i)$
length of median

$A D = \frac{1}{2} \sqrt{2 ({A B}^{2} + {A C}^{2}) - {B C}^{2}}$

$P M = \frac{1}{2} \sqrt{2 ({P Q}^{2} + {P R}^{2}) - {Q R}^{2}}$

$u s i n g (i) & (i i)$

$A D = \frac{1}{2} \sqrt{2 k^{2} ({P Q}^{2} + {P R}^{2}) - {B C}^{2}}$

$\frac{A D}{P M} = k$

$\frac{{A D}^{2}}{{P M}^{2}} = k^{2}$

$\frac{2 k^{2} ({P Q}^{2} + {P R}^{2}) - {B C}^{2}}{2 ({P Q}^{2} + {P R}^{2}) - {Q R}^{2}} = k^{2}$

${2 k}^{2} ({P Q}^{2} + {P R}^{2}) - {B C}^{2} = 2 k^{2} (P Q^{2} + {P R}^{2}) - {Q R}^{2} {(k)}^{2}$

${B C}^{2} = {Q R}^{2} {(k)}^{2}$

$o r \frac{B C}{Q R} = k$

$h e n c e, \frac{A B}{P Q} = \frac{A C}{P R} = \frac{B C}{Q R}$

$∴ △ A B C \sim △ P Q R b y S . S . S s i m i l a r i t y$

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