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Question Number 4505 by Yozzii last updated on 03/Feb/16

Commented by Yozzii last updated on 03/Feb/16

$△ A C D a n d △ D B E a r e e q u i l a t e r a l .$ $F i s t h e m i d p o i n t o f A E a n d G i s t h e$ $m i d p o i n t o f B C . D i s a p o i n t o n t h e$ $l i n e A B . P r o v e t h a t △ F G D i s$ $e q u i l a t e r a l .$ $(T h e d i a g r a m i s o n l y a g u i d e .)$
	Answered by Rasheed Soomro last updated on 05/Feb/16

	$^{∙} L e t c o m m o n v e r t e x o f t r i a n g l e s D = (0, 0)$ $a n d A D a n d D B a r e a l o n g x - a x i s .$ $(A D a n d D B a r e c o l l i n e a r)$ $^{∙} L e t e a c h s i d e o f △ A D C = a u n i t s$ $a n d e a c h s i d e o f △ B D E = b u n i t s$ $^{∙} L e t A i s o n l e f t s i d e o f D a n d B$ $i s o n r i g h t s i d e o f D$ $S o A = (- a, 0) a n d B = (b, 0)$ $I t c a n b e s h o w n e a s i l y$ $t h a t C = (- \frac{a}{2}, \frac{\sqrt{3}}{2} a) a n d E = (\frac{b}{2}, \frac{\sqrt{3}}{2} b)$ $G i s m i d p o i n t o f B C$ $S o G = (\frac{1}{2} (- \frac{a}{2} + b), \frac{1}{2} (\frac{\sqrt{3}}{2} a + 0)) = (\frac{2 b - a}{4}, \frac{\sqrt{3} a}{4})$ $F i s m i d p o i n t o f A E$ $S o F = (\frac{1}{2} (- a + \frac{b}{2}), \frac{1}{2} (\frac{\sqrt{3}}{2} b + 0)) = (\frac{b - 2 a}{4}, \frac{\sqrt{3} b}{4})$ $N o w i n △ F G D$ $F G = \sqrt{{(\frac{2 b - a}{4} - \frac{b - 2 a}{4})}^{2} + {(\frac{\sqrt{3} a}{4} - \frac{\sqrt{3} b}{4})}^{2}}$ $= \sqrt{\frac{{(b + a)}^{2}}{16} + \frac{3 {(a - b)}^{2}}{16}} =$ $= \frac{\sqrt{a^{2} + 2 a b + b^{2} + 3 a^{2} - 6 a b + 3 b^{2}}}{4}$ $= \frac{\sqrt{4 a^{2} - 4 a b + 4 b^{2}}}{4} = \frac{1}{2} \sqrt{a^{2} - a b + b^{2}}$ $F D = \sqrt{{(\frac{b - 2 a}{4} - 0)}^{2} + {(\frac{\sqrt{3} b}{4} - 0)}^{2}}$ $= \sqrt{\frac{b^{2} - 4 a b + 4 a^{2} + 3 b^{2}}{16}} = \frac{1}{2} \sqrt{a^{2} - a b + b^{2}}$ $G D = \sqrt{{(\frac{2 b - a}{4} - 0)}^{2} + {(\frac{\sqrt{3} a}{4} - 0)}^{2}}$ $= \sqrt{\frac{a^{2} - 4 a b + 4 b^{2} + 3 a^{2}}{16}} = \frac{1}{2} \sqrt{a^{2} - a b + b^{2}}$ $∵ F G = F D = G D = \frac{1}{2} \sqrt{a^{2} - a b + b^{2}}$ $∴ △ G F D i s a n EQUILATERAL T r i a n g l e$
	Commented by Yozzii last updated on 04/Feb/16

	$N i c e a p p r o a c h . I s t h e r e a n a n s w e r t h a t$ $u s e s o n l y t h e a x i o m s a n d t h e o r e m s o f$ $E u c l i d e a n g e o m e t r y ?$
	Commented by Rasheed Soomro last updated on 05/Feb/16

	$T h e r e m u s t b e a x i o m a t i c p r o o f, b u t I$ $a m n o t s u c c e s s f u l t o p r o v e y e t .$
	Commented by Yozzii last updated on 06/Feb/16

	$Y e s . T h i s a r e a o f g e o m e t r y i s a l i e n t o$ $m e .$
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