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Question Number 70672 by ajfour last updated on 06/Oct/19

Commented by ajfour last updated on 06/Oct/19

$F i n d s i d e s o f △ A B C . A l s o f i n d$ $t h e s a m e i f i n s t e a d o f E F,$ $D F = 2 .$
	Answered by mind is power last updated on 06/Oct/19

	$x = A D$ $\Rightarrow \frac{2}{x} = \frac{E B}{E B + 3} \Rightarrow 2 E B + 6 = x E B \Rightarrow E B = \frac{6}{x - 2}$ $\Rightarrow \frac{E A}{A B} = \frac{A D}{A C} \Leftrightarrow \frac{3}{3 + \frac{6}{x - 2}} = \frac{x}{x + 5} \Rightarrow \frac{(x - 2)}{x} = \frac{x}{x + 5} \Rightarrow x^{2} = x^{2} + 3 x - 10$ $\Rightarrow x = \frac{10}{3} = A D$ $E B = \frac{18}{7}, B F = \sqrt{{(\frac{18}{7})}^{2} - 4} = \sqrt{\frac{128}{49}} = \frac{8 \sqrt{2}}{7}$ $\frac{B F}{B D} = \frac{2}{A D} \Rightarrow D B = \frac{B F . A D}{2} = \frac{40 \sqrt{2}}{21}$ $F D = \frac{40 \sqrt{2}}{21} - \frac{8 \sqrt{2}}{7} = \frac{16 \sqrt{2}}{7}$
	Commented by ajfour last updated on 06/Oct/19

	$e x c e l l e n t, t h a n k s! b u t E B = \frac{18}{4} .$
	Commented by mind is power last updated on 06/Oct/19

	$y e a h$
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