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 5 of 6: Reflect the orthocenter

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In plain words

Reflection turns an interior altitude distance into a point on the circumcircle, where the arc midpoint gives the largest possible height over $BC$ .

[BCH]=[BCH_1]\le[BCA_1]

Detailed analysis

Let $H$ be the orthocenter and $H_1$ its reflection in $BC$ . Then $H_1$ lies on the circumcircle, and reflection preserves distance to $BC$ , so $[BCH]=[BCH_1]$ . On the arc $BC$ not containing $A$ , $A_1$ is the midpoint and is farthest from the chord $BC$ ; hence $[BCH_1]\le[BCA_1]$ .