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 \(Z\).

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 \(D=\sqrt{B^{2}+E^{2}-2BEcos(\theta)}\)

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)

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 \(A\).

The simple observation that for drawing the line inside a triangle which could be greater than the greatest side must be parallel to the greatest side and hence triangle being a closed figure the required line will not be greater than the largest side. a simple but effective one .. !! :)

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TopNewestWe 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).

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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.

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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 \(A\).

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The simple observation that for drawing the line inside a triangle which could be greater than the greatest side must be parallel to the greatest side and hence triangle being a closed figure the required line will not be greater than the largest side. a simple but effective one .. !! :)

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