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Question Number 207035 by mr W last updated on 04/May/24

	Answered by A5T last updated on 04/May/24

	$B E = \sqrt{6^{2} + 2^{2}} = 2 \sqrt{10} \Rightarrow E F = F B = \sqrt{10}$ $L e t t h e l i n e t h r o u g h F p a r a l l e l t o B C m e e t A B, D C$ $a t H, I r e s p ., t h e n B H = 3 = D I; F H = 1$ $\Rightarrow F I = 5; D G = x \Rightarrow F G = \sqrt{{(3 - x)}^{2} + 5^{2}}$ $\Rightarrow {F G}^{2} = x^{2} - 6 x + 34; E G = \sqrt{4^{2} + x^{2}}$ ${E G}^{2} = {F E}^{2} + {F G}^{2} - 2 F E \times F G c o s 45$ $\Rightarrow 16 + x^{2} = 10 + {F G}^{2} - 2 \sqrt{5} F G$ $\Rightarrow 16 - 10 - 34 + 6 x = - 2 \sqrt{5} F G = 6 x - 28$ $\Rightarrow 6 x - 28 = - 2 \sqrt{5} [\sqrt{{(3 - x)}^{2} + 5^{2}}] \Rightarrow x = \frac{1}{2}$
	Commented by mr W last updated on 04/May/24
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	Answered by mr W last updated on 04/May/24

	Commented by mr W last updated on 04/May/24

	$\tan α = \frac{6}{2} = 3$ $\tan β = \tan (α - 45 °) = \frac{3 - 1}{1 + 3 \times 1} = \frac{1}{2}$ $G H = 5 \tan β = \frac{5}{2}$ $? = D H - G H = 3 - \frac{5}{2} = \frac{1}{2} ✓$
	Answered by HeferH24 last updated on 05/May/24

	$\frac{4 + y}{2 \sqrt{10}} = \frac{\sqrt{10}}{4}$ $16 + 4 y = 20; y = 1$ $\Rightarrow \frac{1}{5} = \frac{x}{3 - x}; x = \frac{1}{2}$
	Commented by HeferH24 last updated on 05/May/24

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