** 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**