Right Triangle

A Triangle with an Angle of 90° ($\pi/2$ radians). The sides $a$, $b$, and $c$ of such a Triangle satisfy the Pythagorean Theorem. The largest side is conventionally denoted $c$ and is called the Hypotenuse.

For any three similar shapes on the sides of a right triangle,

$$A_1+A_2=A_3,$$ (1)

which is equivalent to the Pythagorean Theorem. For a right triangle with sides $a$, $b$, and Hypotenuse $c$, let $r$ be the Inradius. Then

$${\textstyle{1\over 2}}ab={\textstyle{1\over 2}}ra+{\textstyl... ...2}}rb+{\textstyle{1\over 2}}rc={\textstyle{1\over 2}}r(a+b+c).$$ (2)

Solving for $r$ gives

$$r={ab\over a+b+c}.$$ (3)

But any Pythagorean Triple can be written

$$a = m^2-n^2$$ (4)

$$b = 2mn$$ (5)

$$c = m^2+n^2,$$ (6)

so (5) becomes

$$r={(m^2-n^2)2mn\over m^2-n^2+2mn+m^2+n^2} = n(m-n),$$ (7)

which is an Integer when $m$ and $n$ are integers.

Given a right triangle $\Delta ABC$, draw the Altitude $AH$ from the Right Angle $A$. Then the triangles $\Delta AHC$ and $\Delta BHA$ are similar.

In a right triangle, the Midpoint of the Hypotenuse is equidistant from the three Vertices (Dunham 1990). This can be proved as follows. Given $\Delta ABC$, let $M$ be the Midpoint of $AB$ (so that $AM=BM$). Draw $DM\vert\vert CA$, then since $\Delta BDM$ is similar to $\Delta BCA$, it follows that $BD=DC$. Since both $\Delta BDM$ and $\Delta CDM$ are right triangles and the corresponding legs are equal, the Hypotenuses are also equal, so we have $AM=BM=CM$ and the theorem is proved.

See also Acute Triangle, Archimedes' Midpoint Theorem, Brocard Midpoint, Circle-Point Midpoint Theorem, Fermat's Right Triangle Theorem, Isosceles Triangle, Malfatti's Right Triangle Problem, Obtuse Triangle, Pythagorean Triple, Quadrilateral, RAT-Free Set, Triangle

References

Beyer, W. H. (Ed.) CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 121, 1987.

Dunham, W. Journey Through Genius: The Great Theorems of Mathematics. New York: Wiley, pp. 120-121, 1990.