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Apollonius's Theorem

In given figu...

Question

In given figure, D is the mid-point of side BC and AE $⊥$ BC. If $B C = a$ , $A C = b$ , $E D = x$ , $A D = p$ and $A E = h$ , prove that:
$c^{2} = p^{2} - a x + \frac{a^{2}}{4}$

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Solution

i) $D$ is the mid-point of $B C$ and $A E ⊥ B C .$
$I n △ A B E,$ ${A B}^{2} = {A E}^{2} + {B E}^{2} - (1)$ (pythagoras thm)

$△ A D E,$ ${A D}^{2} = {A E}^{2} + {E D}^{2} - (2)$ (pythagoras thm)

$⟹ {A B}^{2} = {A D}^{2} - {E D}^{2} + {B E}^{2}$
$∴ {A B}^{2} = {A D}^{2} - {D E}^{2} + {(B D - D E)}^{2}$
${A B}^{2} = {A D}^{2} - {D E}^{2} + {B D}^{2} + {D E}^{2} - 2 B D \times D E$
$∴ {A B}^{2} = {A D}^{2} - 2 B D \times D E + {B D}^{2}$
$∴ {A B}^{2} = {A D}^{2} - 2 B D \times D E + {B D}^{2}$

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

${A B}^{2} = {A D}^{2} - B C \times D E + \frac{{B C}^{2}}{4}$
$A B = c, A D = p, B C = a, D E = x$
$c^{2} = p^{2} - a x + \frac{a^{2}}{4}$

ii) $C o n s i d e r △ A E C, A D = p, B C = a, A C = b$
${A C}^{2} = {A E}^{2} + {E C}^{2}$
$⟹ {A C}^{2} = {A E}^{2} + {(E D + D C)}^{2} - (1)$

$F r o m △ A E D$
$⟹ {A D}^{2} = {A E}^{2} + {E D}^{2} ⟹ {A E}^{2} = {A D}^{2} - {E D}^{2} - (2)$

$F r o m (1)$
${A C}^{2} = {A D}^{2} - {E D}^{2} + {E D}^{2} + 2 E D . D C + {D C}^{2}$
${A C}^{2} = {A D}^{2} + 2 \times x \times \frac{a}{2} + {(\frac{B C}{2})}^{2} [D C = \frac{B C}{2}]$
$⟹ {A C}^{2} = {A D}^{2} + a x + \frac{a^{2}}{4}$
$A C = b; A D = D$
$⟹ b^{2} = p^{2} + a x + \frac{a^{2}}{4}$

iii) $C o n s i d e r, △ A E C, A C = b, A B = c$
${A C}^{2} = {A E}^{2} + {E C}^{2} - (1)$

$F r o m △ A B E; {A B}^{2} = {A E}^{2} + {B E}^{2} - (2)$

$F r o m △ A E D; {A D}^{2} = {A E}^{2} + {E D}^{2}$

$∴ {A E}^{2} = {A D}^{2} - {E D}^{2} - (3)$

By adding $(1) & (2)$
${A C}^{2} + {A B}^{2} = 2 {A E}^{2} + {E C}^{2} + {B E}^{2}$
$= 2 ({A D}^{2} - {E D}^{2}) + {E C}^{2} + {B E}^{2}$
${A C}^{2} + {A B}^{2} = 2 {A D}^{2} - 2 {E D}^{2} + {E C}^{2} + {B E}^{2}$
$= 2 {A D}^{2} - 2 {E D}^{2} + ({E C}^{2} + {B E}^{2})$
$= 2 {A D}^{2} - 2 {E D}^{2} + {B C}^{2}$

$⟹ b^{2} + c^{2} = 2 p^{2} + \frac{a^{2}}{4}$

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c^{2} = p^{2} - a x + \frac{a^{2}}{4}

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b^{2} = p^{2} + a x + \frac{a^{2}}{4}

b^{2} = p^{2} + a x + \frac{a^{2}}{4}

c^{2} = p^{2} - a x + \frac{a^{2}}{4}

b^{2} + c^{2} = 2 p^{2} + \frac{a^{2}}{2}

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b^{2} + c^{2} = 2 p^{2} + \frac{a^{2}}{2}

In Fig. 7.223, D is the mid-point of side BC and $A E ⊥ B C$ . If BC = a AC=b, AB=c, ED=z, AD = p and AE =h, prove that:

(i) $b^{2} = p^{2} + a x + \frac{a^{2}}{4}$
(ii) $c^{2} = p^{2} + a x + \frac{a^{2}}{4}$
(iii) $b^{2} + c^{2} = 2 p^{2} + \frac{a^{2}}{4}$

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