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Question Number 210172 by efronzo1 last updated on 01/Aug/24

	Answered by a.lgnaoui last updated on 02/Aug/24

	$\frac{\sin 2 a}{BE} = \frac{\cos a}{AB} (∡ AED = 90 - 2 a)$ $AB = \frac{DEcos a}{\sin 2 a} . = \frac{BE}{2 \sin a}$ $Aire (ABC) = Air e (ADFC) + Aire (DBF)$ $A ire (BDE) = Aire (BEF) + Aire (BDF)$ $\Rightarrow$ $Aire (ABC) - (ADFC) =$ $Aire (BDE) - 38$ ${\begin{cases} Aire (ABC) = \frac{1}{2} AB \times AC \sin 2 a \\ Aire (BDE) = \frac{1}{2} BD . BE \cos a \end{cases}$ $AC = BD alors$ $\frac{2 BE . \cos a}{4} - A = \frac{1}{2} BE \cos a - 38$ $Donc Aire (ADFC) = 38$
	Commented by a.lgnaoui last updated on 02/Aug/24

	Answered by mr W last updated on 02/Aug/24

	$s a y A C = B D = b$ $D E = \frac{b}{\tan α}$ $A D = \frac{b}{\tan α \tan 2 α}$ $A r e a (B D E) = \frac{b^{2}}{2 \tan α}$ $A r e a (A B C) = \frac{b \sin 2 α}{2} (b + \frac{b}{\tan α \tan 2 α})$ $= \frac{b^{2}}{2 \tan α} (\tan α \sin 2 α + \cos 2 α)$ $= \frac{b^{2}}{2 \tan α} (2 \sin^{2} α + 1 - 2 \sin^{2} α)$ $= \frac{b^{2}}{2 \tan α} = A r e a (D B E)$ $\Rightarrow A r e a (A D F C) = A r e a (B E F) = 38$
	Answered by A5T last updated on 02/Aug/24

	$\frac{[A B C]}{[B D E]} = \frac{A B \times A C s i n (2 a)}{B D \times D E} = \frac{A B s i n (2 a)}{D E}$ $s i n (2 a) = \frac{D E}{A E}; ∠ B E A = 90 - a = ∠ A B E \Rightarrow A B = A E$ $\Rightarrow \frac{[A B C]}{[B D E]} = \frac{A B s i n (2 a)}{A E s i n (2 a)} = 1$ $\Rightarrow 38 + [D B F] = [A D F C] + [D B F] \Rightarrow [A D F C] = 38$
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