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How do squares of the sides of a triangle, any triangle, relate?

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  • $\begingroup$ It will be great if you draw the two areas $ a \cdot b \cdot \sin C_1$ where $C_1$is the complement. $\endgroup$ – Narasimham May 2 '16 at 0:10
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    $\begingroup$ See this answer. $\endgroup$ – Blue May 2 '16 at 0:20
  • $\begingroup$ So fine, and convincing... The portions around orthocenter simply disappear for the Pythagoras case.. $\endgroup$ – Narasimham May 2 '16 at 10:37
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The general rule that relates the sides of a triangle is called the law of cosines. It regards the following situation, where you know an angle of the triangle and want to relate the side lengths: enter image description here

It is stated as follows: $$C^2=A^2+B^2-2AB\cos(\alpha).$$ Note that this has a very similar form to the Pythagorean theorem - just with an extra adjustment for the angle $\alpha$. In particular, if $\alpha$ is a right angle, this last term is zero, so it reduces to the Pythagorean theorem. The equation also tells us that when $\alpha$ is obtuse, the opposite is longer than it would be in a right triangle and when $\alpha$ is acute, the opposite side is shorter.

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    $\begingroup$ Often opposite side-angle pairs are given "related" symbols, for example if one side/angle is called $A$, then the opposite angle/side is called $\alpha$, $a$ or similar (example: Wikipedia, Solution of triangles). However, you did not follow that practice (in your formula, $\alpha$ and $A$ must be adjacent). Just saying, so that people who are used to such a convention, will not get confused. $\endgroup$ – Jeppe Stig Nielsen May 2 '16 at 9:07
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Sorry for the short answer.

Cosine Rule, whose special case is the Pythagoras theorem.

$$ c^2= a^2 + b^2 - 2 a b \cos C$$

when $C = 90^0. $

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