Right triangle

The diagram for Euclid's proof of the Pythagorean theorem: each smaller square has area equal to the rectangle of corresponding color.

The three sides of a right triangle are related by the Pythagorean theorem, which in modern algebraic notation can be written

${\displaystyle a^{2}+b^{2}=c^{2},}$

where c is the length of the hypotenuse (side opposite the right angle), and a and b are the lengths of the legs (remaining two sides). Pythagorean triples are integer values of a, b, c satisfying this equation. This theorem was proven in antiquity, and is proposition I.47 in Euclid's Elements: "In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle."

As with any triangle, the area is equal to one half the base multiplied by the corresponding height. In a right triangle, if one leg is taken as the base then the other is height, so the area of a right triangle is one half the product of the two legs. As a formula the area T is

${\displaystyle T={\tfrac {1}{2}}ab}$

where a and b are the legs of the triangle.

If the incircle is tangent to the hypotenuse AB at point P, then denoting the semi-perimeter (a + b + c) / 2 as s, we have PA = s − a and PB = s − b, and the area is given by

${\displaystyle T={\text{PA}}\cdot {\text{PB}}=(s-a)(s-b).}$

This formula only applies to right triangles.[3]

Altitude 'f' of a right triangle

If an altitude is drawn from the vertex with the right angle to the hypotenuse then the triangle is divided into two smaller triangles which are both similar to the original and therefore similar to each other. From this:

• The altitude to the hypotenuse is the geometric mean (mean proportional) of the two segments of the hypotenuse.[4]^: 243
• Each leg of the triangle is the mean proportional of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.

In equations,

${\displaystyle \displaystyle f^{2}=de,}$ (this is sometimes known as the right triangle altitude theorem)

${\displaystyle \displaystyle b^{2}=ce,}$

${\displaystyle \displaystyle a^{2}=cd}$

where a, b, c, d, e, f are as shown in the diagram.[5] Thus

${\displaystyle f={\frac {ab}{c}}.}$

Moreover, the altitude to the hypotenuse is related to the legs of the right triangle by[6][7]

${\displaystyle {\frac {1}{a^{2}}}+{\frac {1}{b^{2}}}={\frac {1}{f^{2}}}.}$

For solutions of this equation in integer values of a, b, f, and c, see here.

The altitude from either leg coincides with the other leg. Since these intersect at the right-angled vertex, the right triangle's orthocenter—the intersection of its three altitudes—coincides with the right-angled vertex.

The radius of the incircle of a right triangle with legs a and b and hypotenuse c is

${\displaystyle r={\frac {a+b-c}{2}}={\frac {ab}{a+b+c}}.}$

The radius of the circumcircle is half the length of the hypotenuse,

${\displaystyle R={\frac {c}{2}}.}$

Thus the sum of the circumradius and the inradius is half the sum of the legs:[8]

${\displaystyle R+r={\frac {a+b}{2}}.}$

One of the legs can be expressed in terms of the inradius and the other leg as

${\displaystyle \displaystyle a={\frac {2r(b-r)}{b-2r}}.}$

• ${\displaystyle \displaystyle a^{2}+b^{2}=c^{2}\quad ({\text{Pythagorean theorem}})}$
• ${\displaystyle \displaystyle (s-a)(s-b)=s(s-c)}$
• ${\displaystyle \displaystyle s=2R+r.}$[9]
• ${\displaystyle \displaystyle a^{2}+b^{2}+c^{2}=8R^{2}.}$[10]

• A and B are complementary.[11]
• ${\displaystyle \displaystyle \cos {A}\cos {B}\cos {C}=0.}$[10][12]
• ${\displaystyle \displaystyle \sin ^{2}{A}+\sin ^{2}{B}+\sin ^{2}{C}=2.}$[10][12]
• ${\displaystyle \displaystyle \cos ^{2}{A}+\cos ^{2}{B}+\cos ^{2}{C}=1.}$[12]
• ${\displaystyle \displaystyle \sin {2A}=\sin {2B}=2\sin {A}\sin {B}.}$

• ${\displaystyle \displaystyle T={\frac {ab}{2}}}$
• ${\displaystyle \displaystyle T=r_{a}r_{b}=rr_{c}}$
• ${\displaystyle \displaystyle T=r(2R+r)}$
• ${\displaystyle \displaystyle T={\frac {(2s-c)^{2}-c^{2}}{4}}=s(s-c)}$
• ${\displaystyle T=PA\cdot PB,}$ where P is the tangency point of the incircle at the longest side AB.[13]

• ${\displaystyle \displaystyle r=s-c=(a+b-c)/2}$
• ${\displaystyle \displaystyle r_{a}=s-b=(a-b+c)/2}$
• ${\displaystyle \displaystyle r_{b}=s-a=(-a+b+c)/2}$
• ${\displaystyle \displaystyle r_{c}=s=(a+b+c)/2}$
• ${\displaystyle \displaystyle r_{a}+r_{b}+r_{c}+r=a+b+c}$
• ${\displaystyle \displaystyle r_{a}^{2}+r_{b}^{2}+r_{c}^{2}+r^{2}=a^{2}+b^{2}+c^{2}}$
• ${\displaystyle \displaystyle r={\frac {r_{a}r_{b}}{r_{c}}}.}$[14]

The altitude of a right triangle from its right angle to its hypotenuse is the geometric mean of the lengths of the segments the hypotenuse is split into. Using Pythagoras' theorem on the 3 triangles of sides (p + q, r, s ), (r, p, h ) and (s, h, q ),
${\displaystyle {\begin{aligned}(p+q)^{2}\;\;&=\quad r^{2}\;\;\,+\quad s^{2}\\p^{2}\!\!+\!2pq\!+\!q^{2}&=\overbrace {p^{2}\!\!+\!h^{2}} +\overbrace {h^{2}\!\!+\!q^{2}} \\2pq\quad \;\;\;&=2h^{2}\;\therefore h\!=\!{\sqrt {pq}}\\\end{aligned}}}$

• ${\displaystyle \displaystyle h={\frac {ab}{c}}}$
• ${\displaystyle \displaystyle m_{a}^{2}+m_{b}^{2}+m_{c}^{2}=6R^{2}.}$[8]^{: Prob. 954, p. 26}
• The length of one median is equal to the circumradius.
• The shortest altitude (the one from the vertex with the biggest angle) is the geometric mean of the line segments it divides the opposite (longest) side into. This is the right triangle altitude theorem.

• The triangle can be inscribed in a semicircle, with one side coinciding with the entirety of the diameter (Thales' theorem).
• The circumcenter is the midpoint of the longest side.
• The longest side is a diameter of the circumcircle ${\displaystyle \displaystyle (c=2R).}$
• The circumcircle is tangent to the nine-point circle.[10]
• The orthocenter lies on the circumcircle.[8]
• The distance between the incenter and the orthocenter is equal to ${\displaystyle {\sqrt {2}}r}$.[8]

Let H, G, and A be the harmonic mean, the geometric mean, and the arithmetic mean of two positive numbers a and b with a > b. If a right triangle has legs H and G and hypotenuse A, then[15]

${\displaystyle {\frac {A}{H}}={\frac {A^{2}}{G^{2}}}={\frac {G^{2}}{H^{2}}}=\phi \,}$

and

${\displaystyle {\frac {a}{b}}=\phi ^{3},\,}$

where ${\displaystyle \phi }$ is the golden ratio ${\displaystyle {\tfrac {1+{\sqrt {5}}}{2}}.\,}$ Since the sides of this right triangle are in geometric progression, this is the Kepler triangle.