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Question Number 37220 by tawa tawa last updated on 10/Jun/18

	Answered by ajfour last updated on 01/Jul/18

	$l e t x a x i s b e D E F$ $s u m o f p e r p e n d i c u l a r s o n t a n g e n t$ $(x a x i s n o w) = y_{A} + y_{B} + y_{C}$ $= 3 y_{G}$ $y_{G} b e i n g y c o o r d i n a t e o f c e n t r o i d$ $o f △ A B C (s a m e a s y c o o r d i n a t e$ $o f c i r c u m c e n t r e) = R$ $L e n g t h o f m e d i a n = R + \frac{R}{2} = \frac{3 R}{2}$ $h e n c e y_{A} + y_{B} + y_{C} = 3 y_{G}$ $________________________$ $W e c a n s e e t h a t c e n t r o i d a n d$ $c i r c u m c e n t r e a r e t h e s a m e f o r$ $a n e q u i l a t e r a l t r i a n g l e; y_{G} = R$ $a s c i r c l e t o u c h e s x - a x i s .$ $H e n c e y_{A} + y_{B} + y_{C} = 3 R$ $w h i c h i s t w i c e t h e l e n g t h o f a$ $m e d i a n (= R + R \sin 30 °) = \frac{3 R}{2}$ $________________________$ $= 2 (l e n g t h o f m e d i a n) .$
	Commented by tawa tawa last updated on 11/Jun/18

	$God bless you sir$
	Commented by tawa tawa last updated on 28/Jun/18

	$please sir, can you produce more explanation on the solution for me$ $to understand better . Thanks you sir and God bless you .$
	Commented by tawa tawa last updated on 01/Jul/18

	$?$
	Commented by tawa tawa last updated on 01/Jul/18

	$wow, i really appreciate sir . God bless you sir$
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