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Question Number 112179 by ajfour last updated on 06/Sep/20

Commented by ajfour last updated on 06/Sep/20

$I f A C = B C = a = 5$ $& A D = B E = b = 6$ $f i n d e q u a t i o n o f l i n e s A C a n d B C .$
	Answered by mr W last updated on 06/Sep/20

	Commented by mr W last updated on 06/Sep/20

	$O D = b - p$ $C E = a - \sqrt{p^{2} + q^{2}}$ $a^{2} = {(a - \sqrt{p^{2} + q^{2}})}^{2} + b^{2} - 2 b (a - \sqrt{p^{2} + q^{2}}) \frac{q}{\sqrt{p^{2} + q^{2}}}$ $\Rightarrow (a - \sqrt{p^{2} + q^{2}}) (a - \sqrt{p^{2} + q^{2}} - \frac{2 b q}{\sqrt{p^{2} + q^{2}}}) = a^{2} - b^{2} . . (i)$ $\tan ∠ B = \frac{b - p}{b + q}$ $\cos ∠ B = \frac{a^{2} + b^{2} - {(a - \sqrt{p^{2} + q^{2}})}^{2}}{2 a b}$ $1 + {(\frac{b - p}{b + q})}^{2} = \frac{4 a^{2} b^{2}}{{[a^{2} + b^{2} - {(a - \sqrt{p^{2} + q^{2}})}^{2}]}^{2}}$ $\sqrt{1 + {(\frac{b - p}{b + q})}^{2}} [a^{2} + b^{2} - {(a - \sqrt{p^{2} + q^{2}})}^{2}] = 2 a b . . (i i)$ $p \approx 1.40282$ $q \approx 0.93054$
	Commented by ajfour last updated on 08/Sep/20

	$t h a n k s f o r s o l v i n g, s i r; b u t i b e l i e v e$ $e x a c t a n s w e r m i g h t b e p o s s i b l e$ $s o m e w a y . .$ $J u s t n o w, i c o u l d p r o v e t h a t$ $e i t h e r ∠ A = ∠ B o r ∠ A = \frac{π}{2} + ∠ B$ $w e t a k e t h e f i r s t c a s e h e r e :$ $\Rightarrow l i n e s a r e p e r p e n d i c u l a r$ $T h e n p = \frac{a}{b} (a - \sqrt{b^{2} - a^{2}})$ $a n d q = \frac{\sqrt{b^{2} - a^{2}}}{b} (a - \sqrt{b^{2} - a^{2}})$ $a n d f o r a = 5, b = 6$ $p = \frac{5}{6} (5 - \sqrt{11}) \approx 1.4028$ $q = \frac{\sqrt{11}}{6} (5 - \sqrt{11}) \approx 0.9305$ $(s a m e a s y o u r a n s w e r S i r)$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$
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