Let A 1 , B 1 , C 1 be the tangency points of the intersection of the excircles of a triangle A B C with the sides B C , C A , A B , respectively. Prove that the circumcircles of A B B 1 and A C C 1 meet on a bisector of ∠ B A C .What I thought: Circle ( A B B 1 ) has ω 1 = b ( s − a ) and circle ( A C C 1 ) has v 2 = c ( s − a ). Their radical axis is given by − y c ( s − a ) + z b ( s − a ) which is equation of A angle bisector and we are done.

Question

Let       A    1    ,      B    1    ,      C    1   be the tangency points of the intersection of the excircles of a triangle   A  B  C with the sides   B  C  ,  C  A  ,  A  B  , respectively. Prove that the circumcircles of   A  B      B    1   and   A  C      C    1   meet on a bisector of   ∠  B  A  C  .What I thought: Circle   (  A  B      B    1    ) has       ω    1    =  b  (  s  −  a  ) and circle   (  A  C      C    1    ) has       v    2    =  c  (  s  −  a  ). Their radical axis is given by   −  y  c  (  s  −  a  )  +  z  b  (  s  −  a  ) which is equation of A angle bisector and we are done.

Alannah Yang · Accepted Answer

Let   D be on             B      C        ¯   so that             A      D        ¯   bisects   ∠  B  A  C. Let   φ be the transformation of the plane formed by composing an inversion with center   A and radius   r  =      A    B    ⋅    A    C  , with a reflection through             A      D        ¯  . Notice that   φ  (  B  )  =  C,   φ  (  C  )  =  B,   φ  (            A      D        ¯    )  =            A      D        ¯  . Let also       B    2    =  φ  (      B    1    ),       C    2    =  φ  (      C    1    ). Notice that   φ  (  (  A  B      B    1    )  )  =            C              B        2              ¯  ,   φ  (  (  A  C      C    1    )  )  =            B              C        2              ¯  . We now wish to prove that             A      D        ¯  ,             B              C        2              ¯  ,             C              B        2              ¯   concur. We do this by Ceva on   △  A  B  C.Let   a  =            B      C        ¯  ,   b  =            C      A        ¯  ,   c  =            A      B        ¯  ,   s  =            a      +      b      +      c        2  . We have                    |                              A                      C            2                          ¯                    |                            |                                          C            2                    B                ¯                    |              =                    |                              A                      C            2                          ¯                    |                            |                              A                      C            2                          ¯                    |            −              |                              A          B                ¯                    |              =                    r        2                    s        −        c                                      r          2                          s          −          c                    −                        r          2                b              =      b          b      +      c      −      s        .Likewise,                    |                              C                      B            2                          ¯                    |                            |                                          B            2                    A                ¯                    |              =                    |                              A                      B            2                          ¯                    |            −              |                              A          C                ¯                    |                            |                                          B            2                    A                ¯                    |              =                              r          2                          s          −          b                    −                        r          2                c                            r        2                    s        −        b              =            b      +      c      −      s        c    .And finally,                    |                              B          D                ¯                    |                            |                              D          C                ¯                    |              =      c    b    ,by the Bisector Theorem. Cross-multiplying yields what we wanted to prove.

Let A_1,B_1,C_1 be the tangency points of the intersection of the excircles of a triangle ABC with the sides BC,CA,AB, respectively. Prove that the circumcircles of ABB_1 and ACC_1 meet on a bisector of angle BAC.

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