Altitudes and Orthocenters
By
Cassian Mosha
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. Prove:
Figure 1
Proof:
Also T is the sum of the areas of triangles AHB, AHC and BHC
So |AB||FH|+|AC||EH|+|BC||DH| =2T
Dividing both sides of the latter equation by 2T, and substituting, we get
||AB||FH|/|AB}}CF| + |AC||EH|/|AC||BE| +|BC||DH|/|BC||AD|| =1
Reducing the fractions we get |FH|/|CF|+|EH|/|BE|+|DH|/|AD|=1 something that we wanted.
Now to get the second result it follows that by subtracting the last result above equation from ||CF|/|CF|+|BE|/|BE|+|AD|/|AD|=3, we will get the second required result.QED
If triangle ABC is an obtuse triangle the result above will not hold anymore since point H the orthocener will be lying outside triangle ABC as shown in the sketch below. To make our result above hold we will refer to triangle HBC.
Figure 2