It's a Matter of Proof

by

Patty Wagner

In "Ceva's Achievement," we proved that:

Given a triangle ABC and points d, e, and f on sides AC, BC, and AB respectively, the cevians are concurrent if and only if:Consider triangle ABC and its altitudes, AD, BE, and CF. These segments appear to be concurrent at point H. Applying Ceva's Theorem to this case gives:

Notice the similar triangles that are created by the altitudes:

If we multiply the equations above, we get

Therefore, by Ceva's Theorem, H is a point of concurrency of the altitudes of ΔABC

Given:ΔABC with orthocenter, H. Let points D,E, and F be the feet of the altitudes from A, B, and C respecfully.

Prove:

Proof:

The image on the right provided the hint that enabled me to coneptually understand that the above statement is true.

If we consider each colored triangle separately, we see that their areas are described by 1/2bh. In each triangle, the height is equal to that given in the numerator of the statement: HD for ΔBHC, HE for ΔCHA, and HF for ΔAHB.

The denominators of the statement give the height of the triangle depending on the orientation of the triangle. In the image below, we would describe the area of ΔABC as 1/2(AD)(BC).

Alternatively, we would describe the area of ΔABC in this orientation as 1/2(CF)(AB). (Of course, the area of ΔABC is the same regardless of the orientation of ΔABC.)

Finally, we can see that 1/2(EB)(CA) will also give me the area of ΔABC.

This all suggests that I ought to consider the ratios of the areas of each of the small triangles to the area of ΔABC. Notice that if I add the areas of all the small triangles, it will equal the area of the large triangle. Thus my ratios should add up to one.

Now we test the concept algebraically:

Q.E.D.

Given:ΔABC with orthocenter, H. Let points D,E, and F be the feet of the altitudes from A, B, and C respecfully.

Prove:Since:

Proof:

- AH = AD - HD
- BH = BE - HE
- CH = CF - HF
then,

However, our proof above showed that

Therefore , we have

Q.E.D.