Editing Triangle inequality (section)

===Right triangle===

[[File:Isosceles triangle made of right triangles.svg|thumb|Isosceles triangle with equal sides {{math|{{overline|AB}} {{=}} {{overline|AC}}}} divided into two right triangles by an altitude drawn from one of the two base angles.]]

In the case of right triangles, the triangle inequality specializes to the statement that the hypotenuse is greater than either of the two sides and less than their sum.<ref name=Palmer>
{{cite book |title=Practical mathematics for home study: being the essentials of arithmetic, geometry, algebra and trigonometry |author=Claude Irwin Palmer |url=https://archive.org/details/practicalmathema00palmiala |page=[https://archive.org/details/practicalmathema00palmiala/page/422 422] |publisher=McGraw-Hill |year=1919}}
</ref>

The second part of this theorem is already established above for any side of any triangle. The first part is established using the lower figure. In the figure, consider the right triangle {{mvar|ADC}}. An isosceles triangle {{mvar|ABC}} is constructed with equal sides {{math|''{{overline|AB}}'' {{=}} ''{{overline|AC}}''}}. From the [[triangle postulate]], the angles in the right triangle {{math|ADC}} satisfy:
:<math> \alpha + \gamma = \pi /2 \ . </math>
Likewise, in the isosceles triangle {{mvar|ABC}}, the angles satisfy:
:<math>2\beta + \gamma = \pi \ . </math>
Therefore,
:<math> \alpha = \pi/2 - \gamma ,\ \mathrm{while} \ \beta= \pi/2 - \gamma /2  \ ,</math>
and so, in particular,
:<math>\alpha < \beta \ . </math>
That means side {{mvar|AD}}, which is opposite to angle {{mvar|α}}, is shorter than side {{mvar|AB}}, which is opposite to the larger angle {{mvar|β}}. But {{math|''{{overline|AB}}'' {{=}} ''{{overline|AC}}''}}. Hence:
:<math>\overline{AC} > \overline{AD} \ . </math>
A similar construction shows {{math|''{{overline|AC}}'' > ''{{overline|DC}}''}}, establishing the theorem.

An alternative proof (also based upon the triangle postulate) proceeds by considering three positions for point {{mvar|B}}:<ref name=Zawaira>
{{cite book |title=A primer for mathematics competitions |chapter-url=https://books.google.com/books?id=A21T73sqZ3AC&pg=PA30 |chapter=Lemma 1: In a right-angled triangle the hypotenuse is greater than either of the other two sides |author1=Alexander Zawaira |author2=Gavin Hitchcock |isbn=978-0-19-953988-8 |year=2009 |publisher=Oxford University Press}}
</ref> (i) as depicted (which is to be proved), or (ii) {{mvar|B}} coincident with {{mvar|D}} (which would mean the isosceles triangle had two right angles as base angles plus the vertex angle {{mvar|γ}}, which would violate the [[triangle postulate]]), or lastly, (iii) {{mvar|B}} interior to the right triangle between points {{mvar|A}} and {{mvar|D}} (in which case angle {{mvar|ABC}} is an exterior angle of a right triangle {{mvar|BDC}} and therefore larger than {{math|''π''/2}}, meaning the other base angle of the isosceles triangle also is greater than {{math|''π''/2}} and their sum exceeds {{mvar|π}} in violation of the triangle postulate).

This theorem establishing inequalities is sharpened by [[Pythagoras' theorem]] to the equality that the square of the length of the hypotenuse equals the sum of the squares of the other two sides.