Try this beautiful problem from the PRMO, 2019 based on triangles and internal bisectors.
Let ABC be a triangle and let D be its circumcircle, The internal bisectors of angles A,B and C intersect D at \(A_1,B_1 and C_1\) the internal bisectors of \(A_1,B_1,C_1\) of the triangle \(A_1B_1C_1\) intersect D at \(A_2,B_2,C_2\). If the smallest angle of triangle ABC is 40 find the magnitude of the smallest angle of triangle \(A_2B_2C_2\) in degrees.
Lines
Algebra
Angles
Answer: is 55.
PRMO, 2019, Question 10
Geometry Vol I to IV by Hall and Stevens
angle \(A_1B_1C_1=90 - \frac{ABC}{2}\) angle \(A_1C_1B_1=90-\frac{ACB}{2}\)
angle \(B_1A_1C_1\)=90-\(\frac{BAC}{2}\)
then angle \(A_2B_2C_2=90-\frac{90-\frac{ABC}{2}}{2}\)=45+\(\frac{ABC}{4}\)=55.
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