Show that in a 30°−60°−90° triangle the altitude on the hypotaneuse divides the hypotaneuse into segments whose length has the ratio 1/3. without using trigonometry.


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Question Number 60043 by Kunal12588 last updated on 17/May/19

$S h o w t h a t i n a 30 ° - 60 ° - 90 ° t r i a n g l e t h e$ $a l t i t u d e o n t h e h y p o t a n e u s e d i v i d e s t h e$ $h y p o t a n e u s e i n t o s e g m e n t s w h o s e l e n g t h$ $h a s t h e r a t i o 1 / 3 .$ $w i t h o u t u s i n g t r i g o n o m e t r y .$
	Answered by tanmay last updated on 17/May/19

	$T h e e q n o f h y p o t a n e o u s$ $\frac{x}{a} + \frac{y}{b} = 1$ $b x + a y = a b$ $y = \frac{- b x}{a} + b \to s l o p s e m_{1} = - \frac{b}{a}$ $e q o f s t l i n e ⊥ t o h y p o t a n e o u s a n d p a s s i n g$ $t h r o u g h o r i g i n (0, 0)$ $y = m_{2} x m_{1} \times m_{2} = - 1$ $m_{2} = \frac{- 1}{\frac{- b}{a}} = \frac{a}{b}$ $y = \frac{a}{b} x$ $s o l v e y = \frac{a}{b} x a n d \frac{x}{a} + \frac{y}{b} = 1$ $t o g e t p o i n t C w h i c h d e v i d e A B i n \frac{1}{3} r a t i o$ $\frac{x}{b} = \frac{y}{a} = k \to x = b k y = a k$ $\frac{b k}{a} + \frac{a k}{b} = 1 \to k (a^{2} + b^{2}) = a b k = \frac{a b}{a^{2} + b^{2}}$ $x = \frac{{a b}^{2}}{a^{2} + b^{2}} y = \frac{a^{2} b}{a^{2} + b^{2}}$ $A (0, b) B (a, 0) O (0, 0) a n d C (\frac{{a b}^{2}}{a^{2} + b^{2}}, \frac{a^{2} b}{a^{2} + b^{2}})$ $l e t C d e v i d e s A B \frac{m}{n} r a t i o$ $\frac{{a b}^{2}}{a^{2} + b^{2}} = \frac{m \times 0 + n \times a}{m + n}$ ${m a b}^{2} + {n a b}^{2} = {n a}^{3} + {n a b}^{2}$ $\frac{m}{n} = \frac{a^{3}}{{a b}^{2}} = \frac{a^{2}}{b^{2}}$ $n o w s l o p e o f A B i s m_{1} = \frac{- b}{a}$ $l e t A B m a k e s a n g l e θ w i t h O B$ $s o s l o p e (m_{1}) = \frac{- b}{a} = t a n (180^{o} - 60^{o}) = - \sqrt{3}$ $\frac{b}{a} = \sqrt{3}$ $\frac{m}{n} = \frac{a^{2}}{b^{2}} = \frac{1}{3} p r o v e d$ $◼ ◼ i f w e t a k e θ = 30^{o}$ $s l o p e = \frac{- b}{a} = t a n (180^{o} - 30^{o}) = - \frac{1}{\sqrt{3}}$ $\frac{b}{a} = \frac{1}{\sqrt{3}}$ $t h e n \frac{m}{n} = \frac{a^{2}}{b^{2}} = 3$ $s o r s t i o n \frac{3}{1} a l s o f u l f i l l . . .$ $n o t e T r i a n g l e O A B$ $O A ⊥ O B z n d A B h y p o t a n e o u s$
	Commented by Kunal12588 last updated on 17/May/19

	$s i r y o u s t i l l u s e d t r i g o n o m e t r y (t a n g e n t)$
	Commented by Kunal12588 last updated on 17/May/19

	$b u t i l i k e y o u r w a y s i r (p l s l e t i t b e h e r e)$
	Answered by mr W last updated on 17/May/19

	Commented by mr W last updated on 17/May/19

	$l e t A B = 1, A F = x$ $\Rightarrow B C = C E = 1$ $\Rightarrow C D = x$ $\Rightarrow F E = E D = \sqrt{x^{2} - 1}$ $\Rightarrow A D = 2 + x$ ${A D}^{2} = {A F}^{2} + {F D}^{2}$ $\Rightarrow {(2 + x)}^{2} = x^{2} + {(2 \sqrt{x^{2} - 1})}^{2}$ $\Rightarrow 4 + 4 x + x^{2} = x^{2} + 4 x^{2} - 4$ $\Rightarrow x^{2} - x - 2 = 0$ $\Rightarrow (x + 1) (x - 2) = 0$ $\Rightarrow x = 2$ $\Rightarrow B D = 1 + 2 = 3$ $\Rightarrow A B / B D = 1 / 3 \Rightarrow p r o v e d$
	Commented by Kunal12588 last updated on 17/May/19

	$s i r i t r i e d p l s c h e c k$ $△ A F C i s e q u i l a t e r a l$ $\Rightarrow F C = A C = A B + B C = 2$ $F C = D C$ $\Rightarrow A C = D C$ $\Rightarrow x = 2$ $\frac{A B}{B D} = \frac{1}{x + 1} = \frac{1}{2 + 1} = \frac{1}{3}$
	Commented by mr W last updated on 17/May/19

	$n i c e p a t h s i r!$
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