Question 1195525

Consider any 3 non-collinear points  {{{A}}},{{{B}}},{{{C}}} .

The center of the circle passing through any two given points lies on the perpendicular bisector of the line segment joining those points. This is easy to prove.


Let  {{{O}}}   be the center of any circle passing through the points  {{{A}}} ,{{{B}}}  . 

Join {{{ OA }}} ,  {{{OB}}}   and draw  {{{OD }}}  be perpendicular to {{{ AB}}}  , touching {{{ AB}}}   at {{{ D}}}  .


<a href="https://ibb.co/FDmgZsb"><img src="https://i.ibb.co/FDmgZsb/main-qimg-1002ab23e6b393154bc4873a6427a964.png" alt="main-qimg-1002ab23e6b393154bc4873a6427a964" border="0"></a>


In  Δ  s {{{OAD}}} ,{{{OBD}}} :

{{{OA=OB}}}   (radius) 
{{{OD}}}   is common
∡{{{ODA}}} =∡{{{ODB=90}}} ∘  (By construction).

So,  Δ {{{OAD}}}=Δ {{{OBD}}} . 

Therefore  {{{AD=BD}}} ⟹ {{{OD}}}⊥  bisector of {{{ AB }}}.

Now consider the perpendicular bisectors of line segments {{{ AB}}}  and  {{{AC}}} . 

As the points  {{{A}}}, {{{B}}}  and {{{ C}}}  are not collinear, the perpendicular bisectors of  {{{AB }}} and  {{{AC}}}  are not parallel, and must intersect; two lines can intersect at only one point (let’s call it  {{{O }}}), which must then be the {{{center }}}of the {{{circle}}} passing through the 3 given points (trivial to show that {{{OA=OB=OC }}}).


As the points  {{{A}}}, {{{B}}}  and {{{ C}}}  are non-collinear, they could be the vertices of a {{{triangle}}}. Hence the proof.