Show that the circumscribed circle passes through the middle of the segment determined by center of the incircle and the center of an excircle.

Question

Cortez Hughes · Accepted Answer

Step 1Proof. Note that the incenter, I is determined by the intersection of the internal angle bisectors of the triangle. We will consider the A-excircle, i.e. the excircle whose center, let&#039;s call it X, is determined by the intersection of the external angle bisectors at vertices B and C. Note that X lies on the line AI as well.Step 2Let   ∠  B  A  C  =  α ,   ∠  A  B  C  =  β and   ∠  A  C  B  =  γ . Furthermore, let   A  X  ∩  (  A  B  C  )  =  M , so we need to show that   M  I  =  M  X . We claim that BICX is a cyclic quadrilateral centered at M. Firstly, we have   ∠  B  A  M  =  ∠  C  A  M  =      α    2   , so   M  B  =  M  C . Next, we show that   △  M  C  I is isosceles. Indeed, we have   ∠  M  I  C  =  ∠  I  A  C  +  ∠  A  C  I  =      α    2    +      γ    2    =  ∠  B  A  M  +  ∠  B  C  I  =  ∠  B  C  M  +  ∠  B  C  I  =  ∠  M  C  ISince   M  B  =  M  C  =  M  I , M is the circumcenter of   △  B  C  I , it remains to show that   X  ∈  (  B  C  I  ) . We have   ∠  B  I  C  =  180      °    −      β    2    −      γ    2   . On the other hand,   ∠  B  X  C  =  180      °    −  ∠  C  B  X  −  ∠  B  C  X  =  180      °    −            180              °            −      β        2    −            180              °            −      γ        2    =      β    2    +      γ    2   .Therefore,   ∠  B  I  C  +  ∠  B  X  C  =  180      °   , i.e.   X  ∈  (  B  C  I  ) and (BCI) is centered at M, so   M  I  =  M  X , as desired.

Jordon Haley · Accepted Answer

Step 1Let I is incircle center, O is circumscribed circle center, X is center of excircle that is tangent to edge AB, P is middle of IX segment.  ∠  I  A  B  =            ∠      C      A      B        2   ,   ∠  X  A  B  =            180              °            −      ∠      C      A      B        2    =  90      °    −  ∠  I  A  B ,   ∠  X  A  I  =  ∠  I  A  B  +  ∠  X  A  B  =  90      °   . In the same way,   ∠  X  B  I  =  90      °   . Then XAIB is cyclic quadrilateral with diameter XI and center P, then   P  A  =  P  B . Then P is on perpendicular bisector of AB, as well as O. Then OP is perpendicular bisector of AB.  ∠  A  O  P  =            ∠      A      O      B        2    =  ∠  A  C  B .  ∠  A  P  O  =            ∠      A      P      B        2    =  ∠  A  X  B  =  180      °    −  ∠  A  I  B  =  ∠  I  A  B  +  ∠  I  B  A  =            ∠      C      A      B        2    +            ∠      C      B      A        2    =            180              °            −      ∠      A      C      B        2    =  90      °    −            ∠      A      C      B        2   .  ∠  P  A  O  =  180      °    −  ∠  A  O  P  −  ∠  A  P  O  =  180      °    −  ∠  A  C  B  −  90      °    +            ∠      A      C      B        2    =  90      °    −            ∠      A      C      B        2    =  ∠  A  P  O , therefore   O  A  =  O  P .

Show that the circumscribed circle passes through the middle of

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