Trigonometry/For Enthusiasts/A Generalization of Pythagoras' Theorem

Arbitrary triangle
At any selected angle of a general triangle of sides a, b, c, inscribe an isosceles triangle such that the equal angles at its base θ are the same as the selected angle. Suppose the selected angle θ is opposite the side labeled c. Inscribing the isosceles triangle forms triangle ABD with angle θ opposite side a and with side r along c. A second triangle is formed with angle θ opposite side b and a side with length s along c, as shown in the figure. Tâbit ibn Qorra stated that the sides of the three triangles were related as:
 * $$ a^2 +b^2 =c(r+s) \ . $$

As the angle θ approaches π/2, the base of the isosceles triangle narrows, and lengths r and s overlap less and less. When θ = π/2, ADB becomes a right triangle, r + s = c, and the original Pythagoras' theorem is regained.

One proof observes that triangle ABC has the same angles as triangle ABD, but in opposite order. (The two triangles share the angle at vertex B, both contain the angle θ, and so also have the same third angle by the triangle postulate.) Consequently, ABC is similar to the reflection of ABD, the triangle DBA in the lower panel. Taking the ratio of sides opposite and adjacent to θ,
 * $$\frac{c}{a} = \frac{a}{r} \ .$$

Likewise, for the reflection of the other triangle,
 * $$\frac{c}{b} = \frac{b}{s} \ . $$

Clearing fractions and adding these two relations:
 * $$ cr +cs = a^2 +b^2 \, $$

the required result.

Credits

 * Original content from Wikipedia Pythagorean Theorem