Prove that all altitudes of a triangle are congruent?

[Nightmare225]

Hello, I am a high school student and having a problem with proving the altitudes of a triangle are concurrent. I know that two of them gotta be concurrent because the sides are not parallel. I can't seem to go further, please help. 🙁

 

[FoBoT]

"altitudes of a triangle"

it has been about 25 years since i had geometry, but "altitudes" with respect to triangles isn't ringing a bell

off to google! we go

are these triangles on a treadmill by any chance?

 

[Nightmare225]

I think a diagram would help: http://www.homeschoolmath.net/teaching/g/altitude/three_altitudes.gif

 

[sao123]

Originally posted by: Nightmare225
Hello, I am a high school student and having a problem with proving the altitudes of a triangle are concurrent. I know that two of them gotta be concurrent because the sides are not parallel. I can't seem to go further, please help. 🙁

those are called right angle bi-sections, i believe...
now what is concurrent? or is it congruent?

 

[FoBoT]

http://www.mathwords.com/a/altitude_triangle.htm

i remember "height of a triangle"

 

[Nightmare225]

Originally posted by: sao123

Originally posted by: Nightmare225
Hello, I am a high school student and having a problem with proving the altitudes of a triangle are concurrent. I know that two of them gotta be concurrent because the sides are not parallel. I can't seem to go further, please help. 🙁

those are called right angle bi-sections, i believe...
now what is concurrent? or is it congruent?

That they all intersect at at least one point.

Originally posted by: FoBoT
http://www.mathwords.com/a/altitude_triangle.htm

i remember "height of a triangle"

Yeah, that's it.

 

[FleshLight]

When in doubt, take a derivative.

 

[sao123]

In geometry, three or more lines are said to be concurrent if they intersect at a single point.

In a triangle, four basic types of concurrent lines are altitudes, angle bisectors, medians, and perpendicular bisectors:

In a triangle, altitudes run from each vertex to the point perpendicular on the opposite line. The point where three altitudes meet is the orthocenter.
Angle bisectors are rays running from the bisector of each angle of the triangle. The all meet at the incenter.
Medians connect the vertexes in a triangle to the midpoint of the opposite side. They meet at the centroid.
Perpendicular bisectors are lines running out of the midpoint of each side in a triangle at 90 degree angles. They meet at the circumcenter.

 

[Pugnax]

http://www.cut-the-knot.org/triangle/altitudes.shtml this has a variety of proofs for the problem at hand.

 

[Nightmare225]

Originally posted by: sao123
In geometry, three or more lines are said to be concurrent if they intersect at a single point.

In a triangle, four basic types of concurrent lines are altitudes, angle bisectors, medians, and perpendicular bisectors:

In a triangle, altitudes run from each vertex to the point perpendicular on the opposite line. The point where three altitudes meet is the orthocenter.
Angle bisectors are rays running from the bisector of each angle of the triangle. The all meet at the incenter.
Medians connect the vertexes in a triangle to the midpoint of the opposite side. They meet at the centroid.
Perpendicular bisectors are lines running out of the midpoint of each side in a triangle at 90 degree angles. They meet at the circumcenter.

Yeah, but how do I prove they meet at the orthocenter?

 

[Nightmare225]

Originally posted by: Pugnax
http://www.cut-the-knot.org/triangle/altitudes.shtml this has a variety of proofs for the problem at hand.

Thanks! That's exactly what I needed

# Via the Euler Line

The argument that shows that three points - the circumcenter O, the centroid M, and the orthocenter H - lie on the same line is reversible.

Indeed, in ABC consider the centroid M and the circumcenter O. If they coincide, then so are the corresponding medians and the perpendicular bisectors. In other words, the medians are perpendicular to the sides and, therefore, coincide with the altitudes. The altitudes then intersect at the centroid of the triangle (which is obviously equilateral in this case.)

Assume that the points O and M are distinct. They define a unique line on which we'll consider a point, denoted as H, such that MH = 2*OM with M lying between O and H. Since also AM = 2*MMa, AHM is similar to MaOM. Elements VI.2 implies that lines OMa and AH are parallel. But the former is perpendicular to BC and, therefore, so is the latter. Similarly, BH AC and CH AB.

 

[sao123]

Aha...
The answer

 

[Nightmare225]

Originally posted by: sao123
Aha...
The answer

Again, thank you! These are exactly what I needed. Thank you to all involved! :thumbsup: