The internal and external bisectors of angle A of a triangle A B C meet B C

datomerki8a5yj

datomerki8a5yj

Answered question

2022-05-09

The internal and external bisectors of angle A of a triangle A B C meet B C (or the extension of B C) at X and X respectively, show that the circles A B C and A X X intersect orthogonally. I have shown that the centre of the circle A X X lies on (line) B C, but cannot proceed further. Any advice would be appreciated.

Answer & Explanation

Oswaldo Rosales

Oswaldo Rosales

Beginner2022-05-10Added 16 answers

The center of the circle X A X it's a mid-point of X X because
X A X = 90 .
Let M be this mid-point, O be circumcenter of Δ A B C and let A X is placed between rays A O and A M.

Thus, in the standard notaition:
O A M = O A X + X A M = α 2 B A O + A X M =
= α 2 ( 90 γ ) + β + α 2 = 90
and we are done!

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