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Hooie
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Simple Geometry Question
« on: Jan 16th, 2004, 9:21pm » |
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If a segment bisects a triangle's angle and the side opposite the angle, is the triangle isosceles?
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John_Gaughan
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Re: Simple Geometry Question
« Reply #1 on: Jan 16th, 2004, 10:26pm » |
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Yes. It's been a while since I've done a geometric proof, so here is my reasoning (not a proof).
:: The segment cuts the triangle into two pieces, two new triangles. First, it bisects an angle, so each triangle has an equal angle. The segment bisects the opposide side, so both triangles have an equal side. The segment itself forms a side of each triangle. So you have an angle and two sides that are equal. By some theorem that I don't remember the name of, this means the triangles must be mirror images of each other. Since they are equal, the third sides are also equal. These form two sides of the original triangle. By definition of an isoceles triangle, these two equal sides show that it is isoceles (you could also do so with the two equal angles, they both imply the same thing) ::
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Icarus
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Re: Simple Geometry Question
« Reply #2 on: Jan 18th, 2004, 12:44pm » |
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You remember in geometry, they had these abbreviations for the various triangle theorems, based on what 3 pieces of information were given as congruent for the two triangles? One, AAA, only indicated that the two triangles were similar.
Those with two angles: AAS and ASA implied congruence, since the two angles congruent implies the third one must be as well (at least for nice Euclidean geometry), so the triangles are similar, and the congruent side means the ratio has to be 1.
Then there is SSS: If all three sides are congruent, the triangles must be. It is essentially a requirement for the concept of distance to make sense.
Finally, there are the two 1 angle cases: SAS, and the one that more politely is refered to as SSA, though the reverse might seem more appropriate, for while SAS works (it is often taken as a postulate, rather than proved as a theorem), SSA is the only case that fails.
To see that it fails, draw the known angle, with the adjacent known side, and an infinite ray to represent the other unknown side. The second known side is attached to the other end of the first, but since we dont know at what angle, view it as swinging freely. A triangle is formed when in its swing, its other endpoint contacts the ray. If this side is longer than the minimum needed to reach the ray, its endpoint will cross the ray at two points in the swing, so there are two triangles for SSA conditions, if the third side has length within the right range.
Now, which theorem would you need to prove that the two semi-triangles in this problem are congruent...
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Hooie
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Re: Simple Geometry Question
« Reply #3 on: Jan 21st, 2004, 10:22pm » |
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 Thanks for the help, you guys. It's pretty crazy how much one can forget in three years. Fortunately, I found a geometry texbook I have.
I ended up using the triangle-bisector theorem. If an angle is bisected by a ray, the ray divides the opposite side into segments that have the same ratio as the other two sides of the triangle.
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