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.

Diego Barr

Diego Barr

Answered question

2022-10-12

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.

Answer & Explanation

Alannah Yang

Alannah Yang

Beginner2022-10-13Added 22 answers

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.

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