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Question Number 199374 by cortano12 last updated on 02/Nov/23

Commented by cortano12 last updated on 02/Nov/23

$prove that B$
Commented by MathematicalUser2357 last updated on 03/Nov/23

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Commented by ajfour last updated on 03/Nov/23

$c o n s i d e r a ⊥ d r o p p e d f r o m A o n$ $B C . A H = h, B M = C M = k,$ $M H = b$ ${A B}^{2} + {A C}^{2} = 2 h^{2} + {(k + b)}^{2} + {(k - b)}^{2}$ $= 2 h^{2} + 2 k^{2} + 2 b^{2}$ $n o w h^{2} + b^{2} = {A M}^{2}, k^{2} = {B M}^{2} = {C M}^{2}$ $h e n c e {A B}^{2} + {A C}^{2} = 2 {A M}^{2} + 2 {B M}^{2}$
	Answered by deleteduser1 last updated on 02/Nov/23

	$W L O G, t r a n s l a t e M t o o r i g i n, r o t a t e B M C$ $s u c h t h a t i t c o i n c i d e s w i t h t h e r e a l a x i s$ $T h e n m = 0, b = - c$ $(b - a) (\bar{b} - \bar{a}) + (c - a) (\bar{c} - \bar{a}) = 2 (m - a) (\bar{m} - \bar{a})$ $+ 2 (b - m) (\bar{b} - \bar{m})$ $\Leftrightarrow∣ b ∣^{2} + ∣ a ∣^{2} - a \bar{b} - \bar{a b} + ∣ c ∣^{2} + ∣ a ∣^{2} - a \bar{c} - c \bar{a}$ $= 2 ∣ m ∣^{2} + 2 ∣ a ∣^{2} - 2 a \bar{m} - 2 \bar{a m} + 2 ∣ b ∣^{2} + 2 ∣ m ∣^{2} - 2 m \bar{b}$ $- 2 \bar{m b}$ $\Leftrightarrow 2 ∣ b ∣^{2} + 2 ∣ a ∣^{2} - a (\bar{b} + \bar{c}) - \bar{a} (b + c) = 2 ∣ a ∣^{2} + 2 ∣ b ∣^{2}$ $\Leftrightarrow 0 = 0; H e n c e i n i t i a l e q n m u s t b e t r u e .$
	Answered by som(math1967) last updated on 02/Nov/23

	Answered by mr W last updated on 02/Nov/23

	$B M = C M$ $\cos ∠ B M A = \frac{{A M}^{2} + {B M}^{2} - {A B}^{2}}{2 \times A M \times B M}$ $\cos ∠ A M C = \frac{{A M}^{2} + {C M}^{2} - {A C}^{2}}{2 \times A M \times C M} = - \cos ∠ B M A$ $\frac{{A M}^{2} + {C M}^{2} - {A C}^{2}}{2 \times A M \times C M} = - \frac{{A M}^{2} + {B M}^{2} - {A B}^{2}}{2 \times A M \times B M}$ ${A M}^{2} + {C M}^{2} - {A C}^{2} = - ({A M}^{2} + {B M}^{2} - {A B}^{2})$ $\Rightarrow 2 {A M}^{2} + 2 {B M}^{2} = {A B}^{2} + {A C}^{2}$
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