Area of a Hexagon Formed by the Vertices and Altitude Extensions of a Triangle

T

≈

16.78

2

×

T

≈

33.55

H

≈

33.55

Let ABC be an acute triangle with area

T

. Extend the altitudes of ABC to meet the circumcircle at A', B', and C' corresponding to A, B, and C, respectively. Let the hexagon AC'BA'CB' have area

H

. Then

H=2T

.