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Apollonius's Theorem

The perpendic...

Question

The perpendicular from a on side BC of a $Δ A B C$ intersects BC at D such that DB = 3 CD. Prove that $2 A B^{2} = 2 A C^{2} + B C^{2}$ .

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Solution

$G i v e n △ A B C w i t h A D ⊥ B C .$

$A l s o, D B = 3 C D .$

$L e t B C = x$

$⟹ C D + D B = x$

$C D + 3 C D = x$

$∴ 4 C D = x ⟹ C D = \frac{x}{4}$

$D B = 3 C D = \frac{3 x}{4}$

$B y u s i n g p y t h a g o r a s t h m .$

${(h y p o t e n u s e)}^{2} = {(h t)}^{2} + {(b a s e)}^{2}$

$I n △ A D C, {A C}^{2} = {A D}^{2} + {D C}^{2}$

${A C}^{2} = {A D}^{2} + {(\frac{x}{4})}^{2} - (i)$

$I n △ A D B, {A B}^{2} = {A D}^{2} + {D B}^{2}$

${A B}^{2} = {A D}^{2} + {(\frac{3 x}{4})}^{2} - (i i)$

$S u b t r a c t i n g (i i) - (i)$

${A C}^{2} - {A B}^{2} = {A D}^{2} + \frac{x^{2}}{16} - ({A D}^{2} + \frac{{9 x}^{2}}{16})$

$= {A D}^{2} + \frac{x^{2}}{16} - {A D}^{2} - \frac{{9 x}^{2}}{16}$

$= \frac{x^{2} - {9 x}^{2}}{16} = \frac{- x^{2}}{2}$

$⟹ 2 {A C}^{2} - 2 {A B}^{2} = - x^{2}$

$P u t t i n g, B C = x$

$∴ 2 {A C}^{2} - 2 {A B}^{2} = - {B C}^{2}$

$⟹ 2 {A C}^{2} + {B C}^{2} = 2 {A B}^{2}$

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