Consider a tetrahedron with an equilateral base and two of the other three faces being right triangles with their right angle points meeting. This leaves the last face to be some isosceles triangle. Let the area of the base equal $A$, the area of one of the right triangular faces equal $B$, and the area of the isosceles face equal $C$.
Show a simple proof that $A^2+B^2=C^2$
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Let the base $A$ have sides $a$, the tetrahedron have height $b$, and the remaining side lengths be $c$.
Using Heron's Formula we have $$A^2 = \frac{3a}{2}(\frac{3a}{2}-a)^3 = \frac{3a^4}{16} $$ $$C^2 = \frac{2c+a}{2}(\frac{2c+a}{2}-a)(\frac{2c+a}{2}-c)^2=\frac{4a^2c^2-a^4}{16} $$ $$ B^2 =\frac{a+b+c}{2}(\frac{a+b+c}{2}-a)(\frac{a+b+c}{2}-b)(\frac{a+b+c}{2}-c)=\frac{-a^4+2a^2(b^2+c^2)-(b^2-c^2)^2}{16}$$ As $a,b,c$ are legs of a right triangle with $c$ the hypotenuse we have $a^2+b^2=c^2$ and $b^2=c^2-a^2$. Then substituting in $B^2$ we have $$B^2 = \frac{-a^4+2a^2(b^2+c^2)-(b^2-c^2)^2}{16} = \frac{-a^4+2a^2(c^2-a^2+c^2)-(b^2-(a^2+b^2))^2}{16}=\frac{4a^2c^2-4a^4}{16}$$ Finally we have $$A^2+B^2 =\frac{3a^4}{16}+\frac{4a^2c^2-4a^4}{16}=\frac{4a^2c^2-a^4}{16}=C^2 $$
Let $a$ be a length-side of the base and let $h$ be an altitude to the base of the pyramid.
Thus, we have $$\frac{3a^4}{16}+\frac{a^2h^2}{4}=\frac{a^2\left(h^2+\frac{3a^2}{4}\right)}{4},$$ which is $$A^2+B^2=C^2$$