Problem 2

Let

ABC

be an acute-angled triangle. The internal bisector of angle

A

meets the circumcircle of

ABC

again at

A_1

; points

B_1

and

C_1

are defined similarly. Let

A_0

be the point where the line

AA_1

meets the external bisectors of angles

B

and

C

; points

B_0

and

C_0

are defined similarly. Prove that the area of triangle

A_0B_0C_0

is twice the area of the hexagon

AC_1BA_1CB_1

, and that this area is at least four times the area of triangle

ABC

Step 1 of 6: Recognize the excenters

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In plain words

The three new vertices are not mysterious: they are exactly the three excenters.

A_0=I_A,\quad B_0=I_B,\quad C_0=I_C

Detailed analysis

By definition, $A_0$ lies on the internal bisector of angle $A$ and the external bisectors at $B,C$ ; hence it is the excenter $I_A$ . Cyclically, $B_0=I_B$ and $C_0=I_C$ , so $A_0B_0C_0$ is the excentral triangle.