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Question Number 38032 by behi83417@gmail.com last updated on 20/Jun/18

Commented by behi83417@gmail.com last updated on 21/Jun/18

$i n t r i a n g l e A \overset{▴}{B C} : ∠ A = 72^{∙}, B C = 10 .$ $D a n d E, c a n m o v e o n B C, b u t s u c h$ $t h a t, a l w y e s : A D = A E .$ $1) e v a l u a t e ∠ D A E, w h e n a r e a o f$ $t r i a n g l e D \overset{▴}{A E}, m e e t s m a x i m u m$ $v a l v e .$ $2) i n o n e p o s i t i o n o f : D & E,$ $w e h a v e : C D = E B, A D = A E .$ $f i n d : ∠ D A E a n d a r e a o f$ $A \overset{▴}{D E}, f o r t h i s s p i c i a l c a s e .$ $3) b y c h a n g i n g p o s i t i o n o f^{″} A^{″},$ $w e c a n m a k e t h e e q u a t i o n :$ $C D = D A = A E = E B .$ $n o w e v a l v a t e : ∠ D A E & (\frac{{a r e a}_{D \overset{▴}{A E}}}{{a r e a}_{A \overset{▴}{B C}}}) .$
	Answered by MrW3 last updated on 25/Jun/18

	$l e t ∠ A = α = 72 °, ∠ B = β, ∠ C = γ$ $(1)$ ${l e t}^{'} s a s s u m e β ⩾ γ,$ $\frac{A B}{\sin γ} = \frac{B C}{\sin α}$ $\Rightarrow A B = \frac{\sin γ}{\sin α} \times a = \frac{\sin (α + β) a}{\sin α}$ $m a x . A_{Δ A D E} = A B \times \cos β \times A B \times \sin β$ $= \frac{a^{2} {[\sin (α + β)]}^{2} \sin 2 β}{2 {[\sin α]}^{2}}$ $= \frac{a^{2}}{2} [\cos^{2} β + \frac{\sin^{2} β}{\tan^{2} α}] \sin 2 β$ $(2)$ $β = γ = \frac{π - α}{2}$ $m a x . A_{Δ A D E} = \frac{a^{2}}{2} [\sin^{2} \frac{α}{2} + \frac{\cos^{2} \frac{α}{2}}{\tan^{2} α}] \sin α$ $(3)$ $I {d o n}^{'} t u n d e r s t a n d w h a t i s m e a n t .$
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