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how can we prove that no line can be drawn within a triangle greater than the greatest side

Note by Kartik Umate
3 years, 10 months ago

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We have triangle $$ABC$$ with largest side $$AB$$. We thus also know that $$\angle ACB$$ is the largest angle of triangle $$ABC$$.

Draw a linesegment in the triangle and call the endpoints $$X$$ and $$Y$$. If $$X$$ and/or $$Y$$ are not on the perimeter of triangle $$ABC$$ we can make a longer linesegment by extending $$XY$$ until both points are.

If $$XY$$ is parallel to $$AB$$ then we have that $$ABC$$ and $$XYC$$ are like triangles. Since $$|XC| \leq |AC|$$, we then also have $$|XY| \leq |AB|$$.

If $$XY$$ is not parallel to $$AB$$, suppose that $$X$$ is the point closest to $$AB$$. Otherwise, mirror your image.

Draw a line parallel to $$AB$$ through $$X$$, intersecting triangle $$ABC$$ in a second point $$Z$$.

We have that $$\angle YXZ \leq \angle CAB$$.

By construction, $$\angle YZX = \angle CBA$$.

This gives the following:

$$\angle XYZ = 180 - \angle YXZ - \angle YZX \geq 180 - \angle CAB - \angle CBA = \angle ACB$$

So it follows that

$$\angle XYZ \geq \angle ACB \geq \angle CAB \geq \angle YXZ$$ and

$$\angle XYZ \geq \angle ACB \geq \angle CBA = \angle YZX$$.

So in triangle $$XYZ$$ side $$XZ$$ is the longest.

Thus we have $$|XY| \leq |XZ| \leq |AB|$$ (the last inequality follows from the third paragraph). · 3 years, 10 months ago

I'm not entirely sure but here is my guess:

Given any triangle with one distinct greatest side length $$A$$ and two shorter side lengths $$B$$ and $$C$$. Draw a line $$D$$ from a corner that intersects side length $$B$$ or $$C$$, say that we intersect length $$C$$, the length is split into two lengths and call the length that is furthest from the greatest length $$E$$. Now given a constant angle opposite from the greatest length $$A$$. We will prove that length $$D$$ cannot under any circumstance be greater than length $$A$$ as we will assume that $$E > C$$. In order to prove that $$D < A$$, we will assume that $$D > A$$ in which we will contradict ourselves to prove that this is not possible.

Firstly using the cosine rule we can come up with:

$$A=\sqrt{B^{2}+C^{2}-2BCcos(\theta)}$$ and $$D=\sqrt{B^{2}+E^{2}-2BEcos(\theta)}$$

As we are assuming $$A < D$$

$$\sqrt{B^{2}+C^{2}-2BCcos(\theta)} < \sqrt{B^{2}+E^{2}-2BEcos(\theta)}$$

Let $$2Bcos(\theta)$$ be a constant $$k$$ and rearrange the equation:

$$B^{2}+C^{2}-Ck < B^{2}+E^{2}-Ek$$

$$C(C-k) < E(E-k)$$ where it is obvious that if we know $$C > E$$, hence $$C(C-k) \neq < E(E-k)$$ therefore we contradict ourselves and hence $$A > D$$. (proved)

I hope this helped. · 3 years, 10 months ago

I'm thinking that another possible way could be:

Similar information as above, but let the lengths be vectors, and hence let $$C$$ be broken into two vectors where $$C=E+F$$, the vector $$D$$ intersects $$\angle BA$$ and vector $$C$$. Now assuming that all vectors are $$> 0$$. We will use the method of contradiction to prove that $$D < A$$ by assuming that $$D > A$$:

$$D=B+E$$

$$A=B+E+F$$

$$B+E > B+E+F$$

Hence this is a contradiction as $$F > 0$$

$$\therefore A > D$$ hence the straight line $$D$$ inside the triangle cannot under any circumstance be longer than the greatest side length $$A$$. · 3 years, 10 months ago