In AB^Δ C : m_b ^( 2) + m_c ^( 2) = 5 m_a ^( 2) prove that : A^( ∧) = 90^( °) m


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Question Number 169101 by mnjuly1970 last updated on 24/Apr/22

$I n A \overset{Δ}{B C} : m_{b}^{2} + m_{c}^{2} = 5 m_{a}^{2}$ $p r o v e t h a t : \overset{\land}{A} = 90^{°}$ $m_{a} : (m e d i a n)$
	Answered by mr W last updated on 24/Apr/22

	$m_{a} = \frac{\sqrt{2 b^{2} + 2 c^{2} - a^{2}}}{2}$ $m_{b} = \frac{\sqrt{2 c^{2} + 2 a^{2} - b^{2}}}{2}$ $m_{c} = \frac{\sqrt{2 a^{2} + 2 b^{2} - c^{2}}}{2}$ $m_{b}^{2} + m_{c}^{2} = \frac{2 c^{2} + 2 a^{2} - b^{2} + 2 a^{2} + 2 b^{2} - c^{2}}{4}$ $m_{b}^{2} + m_{c}^{2} = \frac{4 a^{2} + b^{2} + c^{2}}{4}$ $5 m_{a}^{2} = \frac{5 (2 b^{2} + 2 c^{2} - a^{2})}{4} = \frac{4 a^{2} + b^{2} + c^{2}}{4}$ $10 b^{2} + 10 c^{2} - 5 a^{2} = 4 a^{2} + b^{2} + c^{2}$ $b^{2} + c^{2} = a^{2}$ $\Rightarrow r i g h t a n g l e d t r i a n g l e w i t h ∠ A = 90 °$
	Commented by mnjuly1970 last updated on 24/Apr/22

	$t h a n k s a l o t s i r W . . . .$
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