Associated Topics || Dr. Math Home || Search Dr. Math

### Orthic Triangle

```
Date: 04/09/99 at 19:13:57
Subject: Geometry

Dear Sir:

Find the angles of the triangle ABC, which is similar to triangle
A1B1C1, where AA1, BB1, and CC1 are the altitudes of triangle ABC.

I tried to guess the answer and I thought the answer must be an
equilateral triangle.

Yours,
Alazah
```

```
Date: 04/10/99 at 04:19:08
From: Doctor Floor
Subject: Re: Geometry

Hi, Alazah,

Triangle A1B1C1 is usually referred to as the "orthic triangle." The
point where AA1, BB1, and CC1 concur is usually referred to as the
"orthocenter," denoted by H.

When ABC is acute we get the following picture, where I have used

We can see that in the quadrilateral BC'HA', angle A' and angle C' are
both 90 degrees, so BC'HA' is a cyclic quadrilateral. This means that
angles HBA' and HC'A' must be equal, since they cut off the same arc
of the circle circumscribing BC'HA'. In triangle BB'C we see that
angle HBA' = angle B'BC = 90 - angle C (degrees).

So angle HC'A' = 90 - angle C.

In the same way we find angle HC'B' = 90 - angle C.

This gives in triangle A'B'C' that angle C' = 180 - 2*angle C.

So now we have:

angle A' = 180 - 2*angle A
angle B' = 180 - 2*angle B
angle C' = 180 - 2*angle C

If we have angle A = angle A', anlge B = angle B' and angle C = angle
C', then we indeed find an equilateral triangle.

In other cases this is less clear. But, for instance, if angle A =
angle B', angle B = angle C', and angle C = angle A', then we would
have similar triangles too. We find

angle C = 180 - 2*angle A
angle A = 180 - 2*angle B
angle B = 180 - 2*angle C

We find then

angle A = 180 - 2*(180 - 2*angle C)
= 180 - 2*(180 - 2*(180 - 2*angle A))

This simplifies to 9*angle A = 540, so angle A = 60. And again we
have an equilateral triangle.

When ABC is obtuse, let's say angle A > 90 degrees, we find other
angle identities (as we did for acute triangles):

angle A' = 2*angle A - 180
angle B' = 2*angle B
angle C' = 2*angle C

Clearly, here we cannot assume angle A' = angle A.

But again, when angle A = angle B', angle B = angle C' and angle C =
angle A', then we would also have similar triangles. This gives

angle C = 2*angle A - 180
angle A = 2*angle B
angle B = 2*angle C

We find

angle C = 2*(2*angle B) - 180
= 2*(2*(2*angle C) - 180

This results in 7*angle C = 180 , so angle C = 25 5/7.

This gives the triangle:

angle C = 25 5/7
angle B = 2*angle C = 51 3/7
angle A = 2*angle B = 102 6/7

So here we have a second example of a triangle that is similar to its
orthic triangle.

I hope this helped!

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search