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 3 of 6: Double the corresponding areas

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In plain words

Moving the third vertex from the midpoint to the endpoint doubles the altitude and therefore the area.

[IBA_0C]=2[IBA_1C],\quad [ICB_0A]=2[ICB_1A],\quad [IAC_0B]=2[IAC_1B]

Detailed analysis

Because $A_1$ is the midpoint of $IA_0$ , areas of triangles with fixed vertex $B$ and third vertex on line $IA_0$ scale with distance from $I$ . Thus $[IBA_0]=2[IBA_1]$ and $[ICA_0]=2[ICA_1]$ , so $[IBA_0C]=2[IBA_1C]$ . The two cyclic analogues follow identically.