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Question Number 47883 by behi83417@gmail.com last updated on 16/Nov/18

Commented by behi83417@gmail.com last updated on 16/Nov/18

$∡ A = 90^{∙}, c o l o r e d s h a p e s s u c h : A C H G,$ $a r e s q u a r e s a n d p o i n t s : L, . . . a r e$ $s q u a r e c e n t e r s .$ $1 . s h o w t h a t : A F i s t h e a n g u l a r$ $b i s e c t o r o f ∡ B A C .$ $You can't use 'macro parameter character #' in math mode$ $3 . s h o w t h a t, F \overset{△}{K L} i s i s o s c a l e w h e n$ $A \overset{△}{B C} i s i s o s c a l e .$ $4 . a r e a o f F \overset{△}{K L} i n t e r m s o f : a, b, c .$ $(a, b, c : s i d e s o f A \overset{△}{B C}) .$
	Answered by behi83417@gmail.com last updated on 16/Nov/18

	$You can't use 'macro parameter character #' in math mode$ $a c i r c l e p a s s i n g f r o m : A, B, C, a l s o$ $p a s s e s f r o m F . b e c a u s e o f :$ $∡ B A C = ∡ B F C = 90^{∙} . B F = F C = \frac{a}{\sqrt{2}}$ $s o B \overset{a r c}{F} = \overset{a r c}{F C} \Rightarrow B \overset{}{A F} = C \overset{}{A F} . s o$ $A F, i s b i s e c t o r o f B \overset{}{A C} .$ $o r a n o t h e r w a y :$ $F A = x$ $c o s F \overset{}{A B} = \frac{{F A}^{2} + {A B}^{2} - {F B}^{2}}{2 F A . A B} =$ $= \frac{x^{2} + c^{2} - {(\frac{a}{\sqrt{2}})}^{2}}{2 x c} = \frac{2 x^{2} + 2 c^{2} - b^{2} - c^{2}}{4 x c} =$ $= \frac{2 x^{2} + c^{2} - b^{2}}{4 x c} (i)$ $c o s F \overset{}{A C} = \frac{{F A}^{2} + {A C}^{2} - {F C}^{2}}{2 F A . A C} =$ $= \frac{x^{2} + b^{2} - {(\frac{a}{\sqrt{2}})}^{2}}{2 x b} = \frac{2 x^{2} + b^{2} - c^{2}}{4 x b} (i i)$ $(i) - (i i) = \frac{b (2 x^{2} + c^{2} - b^{2}) - c (2 x^{2} + b^{2} - c^{2})}{4 x b c} =$ $= \frac{2 x^{2} (b - c) - (b^{3} - c^{3} - {b c}^{2} + {c b}^{2})}{4 x b c} =$ $= \frac{(b - c) (2 x^{2} - (b^{2} + c^{2} + b c) - b c)}{4 x b c} =$ $= \frac{(b - c) (2 x^{2} - {(b + c)}^{2})}{4 x b c} = 0$ $\Rightarrow ∡ F A B = ∡ F A C \Rightarrow A F, i s b i s e c t o r .$ $[A B C F, i s c y c l i c . s o :$ $x . a = \frac{a}{\sqrt{2}} . b + \frac{a}{\sqrt{2}} . c \Rightarrow x = \frac{b + c}{\sqrt{2}}]$ $You can't use 'macro parameter character #' in math mode$ $t h e a n s w e r i s : n o .$ $S_{F \overset{△}{K L}} = \frac{1}{2} F A . K L = \frac{1}{2} . \frac{b + c}{\sqrt{2}} . \frac{b + c}{\sqrt{2}} = \frac{{(b + c)}^{2}}{4} .$
	Answered by behi83417@gmail.com last updated on 16/Nov/18

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