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Question Number 71423 by ajfour last updated on 15/Oct/19

Commented by ajfour last updated on 15/Oct/19

$M j S S i r . . .$
Commented by mr W last updated on 15/Oct/19

$s u c h t h a t i n c i r c l e e x i s t s,$ $a + c = b + d$
Commented by MJS last updated on 15/Oct/19

$I^{'} ll try . . .$
Commented by ajfour last updated on 15/Oct/19

$g r a n t e d, d = a + c - b$ $f i n d r_{m a x} i n t e r m s o f a, b, c$ $d e a r S i r .$
Commented by ajfour last updated on 15/Oct/19

Commented by ajfour last updated on 16/Oct/19
$let φ=∠AEB Area of quad=(r/2)(a+b+c+d) = r(a+c) [As a+c=b+d] =(1/2)absin 2θ+(1/2)c(a+c−b)sin 2φ ......(i) (φ=∠AEB) AB^2 = a^2 +b^2 −2abcos 2θ = c^2 +(a+c−b)^2 −2c(a+c−b)cos 2φ differentiating we get absin 2θ=c(a+c−b)(sin 2φ)(dφ/dθ) and differentiating (i) (a+c)(dr/dθ)=abcos 2θ +c(a+c−b)(cos 2φ)(dφ/dθ) ⇒ for (dr/dθ)=0 abcos 2θ=−c(a+c−b)cos 2φ ×((absin 2θ)/(c(a+c−b)sin 2φ)) ⇒ tan 2φ=−tan 2θ hence (a+c)r=(1/2)[ab+c(a+c−b)]sin 2θ & AB^2 = a^2 +b^2 −2abcos 2θ = c^2 +(a+c−b)^2 +2c(a+c−b)cos 2θ ⇒ cos 2θ=((a^2 +b^2 −c^2 −(a+c−b)^2 )/(2[ab+c(a+c−b)])) r_(max) = (([ab+c(a+c−b)]sin 2θ)/(2(a+c))) r_(max) =(([ab+c(a+c−b)])/(2(a+c)))(√(1−{((a^2 +b^2 −c^2 −(a+c−b)^2 )/(2[ab+c(a+c−b)]))}^2 )) .$
$l e t ϕ = ∠ A E B$ $A r e a o f q u a d = \frac{r}{2} (a + b + c + d)$ $= r (a + c) [A s a + c = b + d]$ $= \frac{1}{2} a b \sin 2 θ + \frac{1}{2} c (a + c - b) \sin 2 ϕ$ $. . . . . . (i)$ $(ϕ = ∠ A E B)$ ${A B}^{2} = a^{2} + b^{2} - 2 a b \cos 2 θ$ $= c^{2} + {(a + c - b)}^{2} - 2 c (a + c - b) \cos 2 ϕ$ $d i f f e r e n t i a t i n g w e g e t$ $a b \sin 2 θ = c (a + c - b) (\sin 2 ϕ) \frac{d ϕ}{d θ}$ $a n d d i f f e r e n t i a t i n g (i)$ $(a + c) \frac{d r}{d θ} = a b \cos 2 θ$ $+ c (a + c - b) (\cos 2 ϕ) \frac{d ϕ}{d θ} \Rightarrow$ $f o r \frac{d r}{d θ} = 0$ $a b \cos 2 θ = - c (a + c - b) \cos 2 ϕ$ $\times \frac{a b \sin 2 θ}{c (a + c - b) \sin 2 ϕ}$ $\Rightarrow \tan 2 ϕ = - \tan 2 θ$ $h e n c e$ $(a + c) r = \frac{1}{2} [a b + c (a + c - b)] \sin 2 θ$ $&$ ${A B}^{2} = a^{2} + b^{2} - 2 a b \cos 2 θ$ $= c^{2} + {(a + c - b)}^{2} + 2 c (a + c - b) \cos 2 θ$ $\Rightarrow \cos 2 θ = \frac{a^{2} + b^{2} - c^{2} - {(a + c - b)}^{2}}{2 [a b + c (a + c - b)]}$ $r_{m a x} = \frac{[a b + c (a + c - b)] \sin 2 θ}{2 (a + c)}$ $r_{m a x} = \frac{[a b + c (a + c - b)]}{2 (a + c)} \sqrt{1 - {\frac{a^{2} + b^{2} - c^{2} - {(a + c - b)}^{2}}{2 [a b + c (a + c - b)]}}^{2}} .$
Commented by ajfour last updated on 16/Oct/19

$s a y i f a = b = c = d$ $r_{m a x} = \frac{a}{2} \cdot$
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