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Question Number 163384 by amin96 last updated on 06/Jan/22

	Answered by mr W last updated on 06/Jan/22

	Commented by mr W last updated on 06/Jan/22

	$b l u e = A_{Δ E F C} = r e d = A_{Δ A B C} = A_{Δ A D C}$ $\Rightarrow E A / / D F$ $\frac{A B}{E C} = \frac{A F}{F C} = \frac{E D}{D C}$ $\frac{x}{1 + x} = \frac{1}{x}$ $\Rightarrow x^{2} - x - 1 = 0$ $\Rightarrow x = \frac{1 + \sqrt{5}}{2} ◼$
	Commented by amin96 last updated on 06/Jan/22

	$y e s s i r . t h a n k s m r W$
	Commented by amin96 last updated on 06/Jan/22

	Could you please tell me why is DF parallel to AE?
	Commented by mr W last updated on 06/Jan/22

	Commented by mr W last updated on 06/Jan/22

	$[Δ E F C] = [Δ A D C]$ $\Rightarrow [Δ E F D] + [Δ D F C] = [Δ A D F] + [Δ D F C]$ $\Rightarrow [Δ E F D] = [Δ A D F]$ $\Rightarrow \frac{1}{2} DF \times {ED}^{'} = \frac{1}{2} \times DF \times {AF}^{'}$ $\Rightarrow {ED}^{'} = {AF}^{'}$ $\Rightarrow E A / / D F$
	Commented by amin96 last updated on 06/Jan/22

	$V E R Y N I C E$
	Commented by Tawa11 last updated on 06/Jan/22

	$Great sir$
	Answered by mr W last updated on 06/Jan/22

	Commented by mr W last updated on 06/Jan/22

	$a n o t h e r m e t h o d$ $A C = 2 x \cos α$ $\frac{A F}{F C} = \frac{A B}{E C}$ $\frac{A C - F C}{F C} = \frac{A B}{E C}$ $\frac{2 x \cos α}{F C} - 1 = \frac{x}{x + 1}$ $F C = \frac{2 x (x + 1) \cos α}{2 x + 1}$ $b l u e = \frac{1}{2} \times (x + 1) \times \frac{2 x (x + 1) \cos α}{2 x + 1} \times \sin α$ $r e d = \frac{1}{2} \times x \times 2 x \cos α \times \sin α$ $b l u e = r e d$ $\frac{{(x + 1)}^{2}}{2 x + 1} = x$ $x^{2} - x - 1 = 0$ $\Rightarrow x = \frac{1 + \sqrt{5}}{2}$
	Commented by Tawa11 last updated on 07/Jan/22

	$Great sir$
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