I figured this out at school while my teacher was droning on about polygons. I'm sure that it's false or has been discovered already--does anyone know?
Given a right triangle ABC, where angle ABC is the right angle, if you make a point D on AC (the hypotenuse) so that BD and AC are perpendicular, two similar right triangles will be produced, so that ABC, ABD, and BCD are all similar. See this picture (credit to Audi0Slave)
Unfortunately, this theorem is not original. http://scratch.mit.edu/forums/viewtopic … 8#p1143868 Thanks to kayybee for pointing it out.
Last edited by Greenatic (2012-02-18 20:35:55)
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It looks pretty true from the awesome drawing I did, but I'm not sure whether it's already been discovered or whether it actually is true. Pretty cool, though.
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Darn, I haven't got any paper to try it out :l
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laptop97 wrote:
Perpendicular or angle bisector I think. Can't remember which one though.
It's the uhhh
Altitude?
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veggieman001 wrote:
laptop97 wrote:
Perpendicular or angle bisector I think. Can't remember which one though.
It's the uhhh
Altitude?
The altitude is what is splitting the triangle from the right angle (vertex) to the hypotenuse (base) but the defention of a perpindicular bisector or angle bisector require more info then that. I think you have to know that some sides are congruent. This seems to fit more with the perpendicular bisector but I don't know for sure
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laptop97 wrote:
veggieman001 wrote:
laptop97 wrote:
Perpendicular or angle bisector I think. Can't remember which one though.
It's the uhhh
Altitude?The altitude is what is splitting the triangle from the right angle (vertex) to the hypotenuse (base) but the defention of a perpindicular bisector or angle bisector require more info then that. I think you have to know that some sides are congruent. This seems to fit more with the perpendicular bisector but I don't know for sure
But it literally is an altitude, because it goes from the vertex to base like you said, and that's all the info that's given. It could be both a perpendicular bisector and an altitude, but that would make it like really limited and it seems to work with just an altitude.
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laptop97 wrote:
If it is an isoceles right triangle where the legs are congruent (obviously) and the line goes from the right angle to the hypotenuse and is perpindicular, they will be congruent right triangles.
Yes, that is true and it's a theorem. But he's saying that in any right triangle, the altitude splits it so it forms two triangles that are similar to the original.
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veggieman001 wrote:
laptop97 wrote:
If it is an isoceles right triangle where the legs are congruent (obviously) and the line goes from the right angle to the hypotenuse and is perpindicular, they will be congruent right triangles.
Yes, that is true and it's a theorem. But he's saying that in any right triangle, the altitude splits it so it forms two triangles that are similar to the original.
So in an isoceles triangle they would be similar to the original and the smaller ones would be congruent to each other. I'm going to try scalene next.
Edit: Nevermind; I'm not going to do math on the weekend
Last edited by laptop97 (2012-02-18 11:49:28)
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laptop97 wrote:
Perpendicular or angle bisector I think. Can't remember which one though.
Neither. Perpendicular bisector requires that the two line segments produced by bisecting are congruent, which would only be true on an isosceles right triangle. Angle bisector requires that the two angles produced by bisecting are congruent, but that is also only true on an isosceles right triangle.
As someone said upthread, altitude might be important. You could rephrase it to say "The altitude of any right triangle will produce two right triangles similar to the orginial," which sounds a lot clearer.
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Greenatic wrote:
laptop97 wrote:
Perpendicular or angle bisector I think. Can't remember which one though.
Neither. Perpendicular bisector requires that the two line segments produced by bisecting are congruent, which would only be true on an isosceles right triangle. Angle bisector requires that the two angles produced by bisecting are congruent, but that is also only true on an isosceles right triangle.
As someone said upthread, altitude might be important. You could rephrase it to say "The altitude of any right triangle will produce two right triangles similar to the orginial," which sounds a lot clearer.
Yeah I realized that after a while, look at the posts after that.
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laptop97 wrote:
veggieman001 wrote:
laptop97 wrote:
If it is an isoceles right triangle where the legs are congruent (obviously) and the line goes from the right angle to the hypotenuse and is perpindicular, they will be congruent right triangles.
Yes, that is true and it's a theorem. But he's saying that in any right triangle, the altitude splits it so it forms two triangles that are similar to the original.
So in an isoceles triangle they would be similar to the original and the smaller ones would be congruent to each other. I'm going to try scalene next.
Edit: Nevermind; I'm not going to do math on the weekend
I tried 30-60-90 and 10-80-90 triangles (referring to the angles), and it appeared to be true.
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laptop97 wrote:
Is there an equation for finding the altitude or do you have to use a protractor and or compass? I looked through 5-6 youtube videos and a wikipedia article, but they all look they just assumed it was perpendicular
The altitude has to be perpendicular.
Wikipedia wrote:
Relation among altitudes of right triangleIn a right triangle the three altitudes α, β, η (the first two of which coincide with the legs b and a respectively) are related according to
(1/(α^2) ) + (1/(β^2) ) = (1/(η^2) )
Last edited by Greenatic (2012-02-18 12:33:33)
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This is false. I tested it, and if you halve the 2 triangles I got, you could get 4 triangles, not 2 more (which I assume makes 3 triangles).
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ABC can't be a right angle if AB is the hypotenuse
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rdococ wrote:
This is false. I tested it, and if you halve the 2 triangles I got, you could get 4 triangles, not 2 more (which I assume makes 3 triangles).
Wait, you're saying you have a counterexample--one right triangle, that when divided by its altitude, makes more than two more triangles? I don't think I understand you--could you post a screenshot of what you're doing?
Audi0Slave wrote:
ABC can't be a right angle if AB is the hypotenuse
Wait, you're right. Whoops. I think you still get the idea though.
Last edited by Greenatic (2012-02-18 14:44:46)
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rdococ wrote:
This is false. I tested it, and if you halve the 2 triangles I got, you could get 4 triangles, not 2 more (which I assume makes 3 triangles).
Could you show an image?
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Greenatic wrote:
Given a right triangle ABC, where angle ABC is the right angle, if you make a point D on AB (the hypotenuse).
http://i44.tinypic.com/1e2ps1.png
Totally.
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So like this? http://i41.tinypic.com/jpifiq.jpg
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veggieman001 wrote:
rdococ wrote:
This is false. I tested it, and if you halve the 2 triangles I got, you could get 4 triangles, not 2 more (which I assume makes 3 triangles).
Could you show an image?
It shows that adding a D point only allows for 2 triangles unless you halve them (which should be ignored in this discussion).
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This doesn't work that well considering the altitude has to be an angle bisector.
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rdococ wrote:
veggieman001 wrote:
rdococ wrote:
This is false. I tested it, and if you halve the 2 triangles I got, you could get 4 triangles, not 2 more (which I assume makes 3 triangles).
Could you show an image?
http://img808.imageshack.us/img808/2432 … xample.png
It shows that adding a D point only allows for 2 triangles unless you halve them (which should be ignored in this discussion).
Uh, I don't think you get it. Look at Audi0Slave's picture
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veggieman001 wrote:
rdococ wrote:
veggieman001 wrote:
Could you show an image?http://img808.imageshack.us/img808/2432 … xample.png
It shows that adding a D point only allows for 2 triangles unless you halve them (which should be ignored in this discussion).Uh, I don't think you get it. Look at Audi0Slave's picture
I see atleast 4 triangles in that picture. Without the little ones, there's just two.
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