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 4 of 6: Tile the two regions

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In plain words

The same pieces are seen at two scales: every piece on the excentral side is twice its counterpart in the hexagon.

[A_0B_0C_0]=2[AC_1BA_1CB_1]

Detailed analysis

The three quadrilaterals on the left partition the excentral triangle $A_0B_0C_0$ (with the orthic triangle $ABC$ inside it), while the corresponding three quadrilaterals on the right partition the hexagon $AC_1BA_1CB_1$ . Summing the three equalities from Step 3 therefore gives $[A_0B_0C_0]=2[AC_1BA_1CB_1]$ .