please i need it urgently show that the midpoint of the hypotenuse of a right triangle is equidistant from its vertices


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Question Number 78427 by Chi Mes Try last updated on 17/Jan/20

$p l e a s e i n e e d i t u r g e n t l y$ $s h o w t h a t t h e m i d p o i n t o f t h e h y p o t e n u s e$ $o f a r i g h t t r i a n g l e i s e q u i d i s t a n t f r o m i t s v e r t i c e s$
	Answered by MJS last updated on 17/Jan/20

	$easy :$ $any right angled triangle is half a rectangle .$ $any rectangle has a circumcircle with center$ $at the intersection of the diagonals \Rightarrow$ $\Rightarrow the vertices have the same distance from$ $the center$
	Answered by $@ty@m123 last updated on 17/Jan/20

	$L e t △ A B C i s r i g h t a n g l e d a t B .$ $L e t D b e t h e m i d p o i n t o f A C .$ $D r a w D E ⊥ B C \Rightarrow D E ∥ A B .$ $C o n v e r s e o f m i d p o i n t t h e o r e m$ $i m p l i e s t h a t E i s m i d p o i n t o f B C .$ $N o w i n △ D B E,$ ${D B}^{2} = {D E}^{2} + {E B}^{2}$ ${D B}^{2} = {(\frac{A B}{2})}^{2} + {(\frac{B C}{2})}^{2}$ ${D B}^{2} = (\frac{{A B}^{2} + {B C}^{2}}{4}) = \frac{{A C}^{2}}{4}$ $\Rightarrow D B = \frac{A C}{2}$ $\Rightarrow D B = A D = D C$ $H e n c e t h e r e s u l t .$
	Commented by $@ty@m123 last updated on 17/Jan/20

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