**by**

**Behnaz
Rouhani**

Given that ABC is an acute triangle, construct the orthocenter H. Let Points D, E, and F be the feet of the perpendiculars from A, B, and C respectively.

Prove that:

**Case I:**

**Case II:**

We should consider what happens if ABC is an obtuse
triangle, and see if the above relationships still hold.

First, we would construct the orthocenter H of triangle ABC, and let the points D, E, and F be the feet of the perpendiculars from A, B, and C respectively.

**Case I:**

We know that

area (ABC) = area (HAB) + area (HAC) + area (HBC)

Now, divide both sides of the above equation by A, to get:

After simplifying the fractions we get:

or

**Case II:**

To find the validity of the second case we know that:

Rearrangement of terms gives

When we have an obtuse triangle the orthocenter will be outside the triangle. In the figure below the orthocenter of triangle ABC is constructed. Then feet of the perpendiculars from vertices A, B, and C are points D, E, and F respectively.

**Case I:**

In this case

area (BCH) = area (ABC) + area (AHB) + area (CAH)

After simplifying fractions we get:

As we can see this is different from equation (1) which we proved earlier.

**Case II:**

To prove the second relationship, we know that:

Do you see any similarities between the relationships for the acute and obtuse triangles?

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