What theorem state that the sum of the two sides of a triangle is greater than or equal to the third side?

Answer

Verified

Hint: Use the fact that in a triangle, the larger angle has a larger opposite side. Extend AC to point D such that CD = BC. Join BD. Observe in triangle ADB $\angle ABD>\angle ADB$. Apply the above-mentioned theorem and hence prove the result.

Complete step-by-step answer:

Given: A triangle ABC.To prove: $AC+BC>AB$Construction: Extend AC to a point D such that CD = CB. Join BD.Proof:In triangle BCD, we haveBC=DCHence $\angle CDB=\angle CBD$ (because equal sides of a triangle have equal opposite angles)Now since $\angle ABD>\angle CBD$, we have$\angle ABD>\angle CDB$.Now in triangle ABD, we have$\angle ABD>\angle CDB$Hence we have $AD>AB$( because the side opposite to a larger angle is longer).Now we haveAD = AC+CDSince CD = BC, we haveAD = AC+BC.Hence $AD>AB\Rightarrow AC+BC>AB$Hence the sum of two sides of a triangle is larger than the third side.Note: [1] The above inequality is strict, i.e. We cannot have a triangle in which the sum of sides is even equal to the third side.[2] The above theorem is necessary for the existence of a triangle and is also sufficient for the existence of the triangle, i.e. if the sum of any two sides is greater than the third side, then the triangle with the given sides exists and if there exists a triangle then the sum of its any two sides is always greater than the third side.

Perhaps not as directly-trigonometric as you want (@Eugen gave the answer I would've given in that regard), but ...

Heron's Formula. If $T$ is the area of the triangle with side-lengths $a$, $b$, $c$, then $$\frac{16\;T^2}{a+b+c} = (-a+b+c)(a-b+c)(a+b-c) \tag{$\star$}$$

Note that each factor on the right-hand side corresponds to an aspect of the Triangle Inequality. For non-degenerate triangles ($T>0$), the left-hand side is strictly positive, which implies that the number of negative factors on the right must be even; but, one readily determines that this number of factors cannot be two, so it must be zero, which is to say: all three Triangle Inequalities must hold. (I'll leave it to the reader to consider the degenerate case ($T=0$).)

Another way to think about this is:

Three lengths form a triangle if and only if Heron calculates a real area ($T$) from them. That is, Heron's formula not only computes a triangle's area, it determines a potential triangle's viability.

FYI: Menger's Theorem characterizing when six lengths form a tetrahedron works similarly: (1) Heron must calculate four real face areas, and (2) the Cayley-Menger determinant must calculate a real volume.

The given statement is False

Justification

The sum of the lengths of any two sides of a triangle is always greater than the length of the third side.

Triangle Inequality Theorem Proof

The triangle inequality theorem describes the relationship between the three sides of a triangle. According to this theorem, for any triangle, the sum of lengths of two sides is always greater than the third side. In other words, this theorem specifies that the shortest distance between two distinct points is always a straight line.

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