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Apollonius's Theorem

The following...

Question

The following figure shows a triangle $A B C$ in which $A D$ is a median and $A E ⊥ B C$ .
Prove that: $2 A B^{2} + 2 A C^{2} = 4 A D^{2} + B C^{2}$

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Solution

In $△ A E B,$
${A B}^{2} = {A E}^{2} + {E B}^{2}$ ..... $(1)$

In $△ A E D,$
${A D}^{2} = {A E}^{2} + {E D}^{2}$ ...... $(2)$

In $△ A E C,$
${A C}^{2} = {A E}^{2} + {E C}^{2}$ ...... $(3)$

Adding $(1)$ and $(3)$ , we get
${A B}^{2} + {A C}^{2} = {2 A E}^{2} + {E C}^{2} + {E B}^{2}$
But ${E C}^{2} + {E B}^{2} = {C B}^{2} - 2 E C . E B$
$∴ {A B}^{2} + {A C}^{2} = {2 A E}^{2} + {C B}^{2} - 2 E C . E B$ ...... $(4)$

From $(2),$ ${A E}^{2} = {A D}^{2} - {E D}^{2}$ ...... $(5)$
Substituting $(5)$ in $(4),$ we get
${A B}^{2} + {A C}^{2} = 2 {A D}^{2} - 2 {E D}^{2} + {C B}^{2} - 2 E C . E B$

$D E = D B - E B$
${D E}^{2} = {D B}^{2} + {E B}^{2} - 2 D B . B E$ --(6)

Substituting (6) in (4)
${A B}^{2} + {A C}^{2} = 2 {A D}^{2} - 2 {D B}^{2} - 2 {E B}^{2} + 4 D B . B E - 2 E C . E B + {C B}^{2} = 2 {A D}^{2} - 2 {D B}^{2} - 2 {E B}^{2} + 2 E B (2 D B - E C) + {C B}^{2} = 2 {A D}^{2} - 2 {D B}^{2} - 2 {E B}^{2} + 2 E B (B C - E C) + {C B}^{2} [C D = B D] = 2 {A D}^{2} - 2 {D B}^{2} - 2 {E B}^{2} + 2 E B (E B) + {C B}^{2} = 2 {A D}^{2} - 2 {D B}^{2} + {C B}^{2}$

But $D B = B C / 2$
${A B}^{2} + {A C}^{2} = 2 {A D}^{2} - \frac{2}{4} {C B}^{2} + {C B}^{2} {2 A B}^{2} + 2 {A C}^{2} = 4 {A D}^{2} + {B C}^{2}$
Hence proved.

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