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 2 of 6: Use the incenter–excenter midpoint lemma

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In plain words

Each arc midpoint is a balance point between the incenter and the matching excenter.

A_1\in IA_0,\qquad A_1I=A_1A_0

Detailed analysis

Let $I$ be the incenter. Angle chasing gives $\angle A_1BI=\angle A_1IB=(A+B)/2$ , hence $A_1B=A_1I$ ; symmetry gives $A_1C=A_1I$ . The analogous external-bisector chase gives $A_1A_0=A_1B$ . Since $I,A_1,A_0$ are collinear, $A_1$ is the midpoint of $IA_0$ . Similarly, $B_1$ and $C_1$ are the midpoints of $IB_0$ and $IC_0$ .