Altitudes and Orthocenters

Rozina Essani


12. Given triangle ABC. Construct the Orthocenter H. Let points D, E, and F be the feet of the perpendiculars from A, B, and C respectfully.







Let us first draw triangle ABC and the altitudes of the vertices. The altitude of  A meets BC at D,  B meets AC at E and C meets AB at F. The point of intersection of the three altitudes is the orthocenter H.

We know that the area of a triangle is (1/2) x base x height.



Area of ABC ===

Area(ABC) = area(BHC) + area (AHB) + area(AHC)

 ===  +  +

1 =

Hence we have proved the first part. Now lets look at the second part.


From the GSP illustration we know that

AH = AD – HD

BH = BE – HE

CH = CF – HF

Now substituting into our original equation we get


from the proof of the first part we have  = 1



Hence .


Let us apply lengths to the segments in triangle ABC and check to see if these equations hold.



Will the same apply to an obtuse triangle?




As we can see by changing to the triangle around to create obtuse triangles that each figure is missing one of three components to the orthocenter. This will not allow for the equations to work on obtuse triangles.