WOLFRAM|DEMONSTRATIONS PROJECT

Bisection of Segments by the Sides of a Triangle

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HA'
≈
1.40
HB'
≈
2.48
HC'
≈
0.84
A'A''
≈
1.40
B'B''
≈
2.48
C'C''
≈
0.84
Let ABC be a triangle with orthocenter H, and let the feet of the altitudes be A', B' and C'. Let the extensions of the altitudes intersect the circumcircle a second time at A'', B'', and C''. Then HA' = A' A'', HB' = B' B'', and HC' = C' C''.