I would like suggestions if possible, other than the really sloppy picture, I'll edit that once my dad gets me Microsoft office. I got a snip of the shape, and edited it as best as I could.
The theorem
For an oblique triangular pyramid if all 3 faces form 90 degrees at one point, then each of the areas of the faces that share the vertex, squared, will be equal to the face opposite of the vertex, squared.
The approach
In order to solve this it is best to prove this using the side lengths. All 3 sides that meet at the intersection will be called $a$, $b$, and $c$. Then using these sides we will find the areas of each face, which will be name $A_1$, $A_2$, and $A_3$, in terms of a, b, and c.
Then we'll use the theorem to find the "expected" area of the opposite face ($A_4$) to the intersection.
From there will use the Pythagorean theorem since the faces $A_1$, $A_2$, and $A_3$ form 90 degrees at one point to find the sides of their hypotenuses.
Using the hypotenuses, they will form the sides of the face $A_4$. Since we know only the sides of $A_4$ we will have to using Heron's formula.
If we continue simplifying we should have the expected area $A_4$ be equal to the actual area of $A_4$, thus proving the theorem.
Solving the theorem
First we will solve for the "expected area" of the theorem. According to the theorem described on the top (1)${A_1}^2+{A_2}^2+{A_3}^2={A_4}^2$. Using the sides $a$, $b$, and $c$, we can solve the areas for each face, using the area of triangles.
(2): Area=1/2(base)(height)
We substitute the sides, each being both a base and a height shown down below.
(3):$A_1=1/2{ab}$
(4):$A_2=1/2{bc}$
(5):$A_3=1/2{ac}$
Using (1) we substitute the values to get:
(6):${A_4}^2=1/4{ab}^2+1/4{bc}^2+1/4{ac}^2$
By solving for $A_4$ we end up getting:
(7):${A_4}=\sqrt{1/4{(ab)}^2+1/4{(bc)}^2+1/4{(ac)}^2}$
This is our "expected value" for proving the theorem. The next part would be solving for the actual area of $A_4$.
From the picture above we can tell that all the hypotenuses of $A_1$, $A_2$, and $A_3$, we have all the sides of $A_4$, since they all have right angles,by using the 2-d Pythagorean theorem.
(8): (Hypotenuse)=(Base)$^2$+(Height)$^2$
Solving for the hypotenuses by square rooting both sides we get:
(9): Hypotenuse of $A_1$ is $\sqrt{a^2+b^2}$
(10): Hypotenuse of $A_2$ is $\sqrt{b^2+c^2}$
(11): Hypotenuse of $A_3$ is $\sqrt{a^2+c^2}$
Now that we know all the sides of $A_4$ we have to solve for its actual area. Since we only know the sides of $A_4$ we can only solve this using Heron's formula.
Using Heron's formula, first we must take the semi-perimeter:
(12): $s=\frac{S_1+S_2+S_3}{2}$
Where s is the semi-perimeter and the subs of Capital S's are all the side lengths. From there we take s, and create:
(13):$\sqrt{s(s-S_1)(s-S_2)(s-S_3)}$
Now if you substitute (9), (10), and (11), we get:
(14):s=$\frac{\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}}{2}$
Than substituting (14) for s, and (9),(10), and (11) for the side lengths we get.
(15):$\sqrt{{\frac{\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}}{2}}{{(\frac{\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}}{2}}-{\sqrt{a^2+b^2}})(\frac{\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}}{2}-\sqrt{b^2+c^2})(\frac{\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}}{2}-\sqrt{a^2+c^2})}}$
This as a whole can be simplified into (16):$\sqrt{(\frac{\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}}{2})(\frac{-\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}}{2})(\frac{\sqrt{a^2+b^2}-\sqrt{b^2+c^2}+\sqrt{a^2+c^2}}{2})(\frac{\sqrt{a^2+b^2}+\sqrt{b^2+c^2}-\sqrt{a^2+c^2}}{2})}$
Since the equation is huge, we will separate the first two parentheses and multiply it out, ignoring the square root of the entire equation for an instant: (17):$(\frac{\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}}{2})(\frac{-\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}}{2})$
Multiplying the the brackets result in: (18):$\frac{(-{\sqrt{a^2+b^2}}^2+{\sqrt{a^2+b^2}\sqrt{b^2+c^2}}+{\sqrt{a^2+b^2}\sqrt{a^2+c^2}}-{\sqrt{b^2+c^2}\sqrt{a^2+b^2}}+{\sqrt{b^2+c^2}}^2+{\sqrt{b^2+c^2}\sqrt{a^2+c^2}}-{\sqrt{a^2+b^2}\sqrt{a^2+c^2}}+{\sqrt{b^2+c^2}\sqrt{a^2+c^2}}+{\sqrt{a^2+c^2}}^2)}{4}$
Cancelling out most terms we get: (19):$\frac{-a^2-b^2+b^2+c^2+c^2+a^2+{\sqrt{b^2+c^2}\sqrt{a^2+c^2}}}{4}$
Further simplification results in: (20):$1/2{c^2}+1/2{\sqrt{b^2+c^2}\sqrt{a^2+c^2}}$
However we must tend to the other two brackets, once again we ignore the square root of all the parenthesis:
(21):$(\frac{\sqrt{a^2+b^2}-\sqrt{b^2+c^2}+\sqrt{a^2+c^2}}{2})(\frac{\sqrt{a^2+b^2}+\sqrt{b^2+c^2}-\sqrt{a^2+c^2}}{2})$
Multiplying this out we end up getting:
(22):$\frac{({\sqrt{a^2+b^2}}^2+{\sqrt{a^2+b^2}\sqrt{b^2+c^2}}-{\sqrt{a^2+b^2}\sqrt{a^2+c^2}}-{\sqrt{b^2+c^2}\sqrt{a^2+b^2}}-{\sqrt{b^2+c^2}}^2+{\sqrt{b^2+c^2}\sqrt{a^2+c^2}}+{\sqrt{a^2+b^2}\sqrt{a^2+c^2}}+{\sqrt{b^2+c^2}\sqrt{a^2+c^2}}-{\sqrt{a^2+c^2}}^2)}{4}$
Once again we cancel out the terms to get:
(23):$\frac{a^2+b^2-b^2-c^2-c^2-a^2+{\sqrt{b^2+c^2}\sqrt{a^2+c^2}}}{4}$
We end up with:
(24):$-1/2{c^2}+1/2{\sqrt{b^2+c^2}\sqrt{a^2+c^2}}$
Now we must take (20), and multiply it by (24):
(25):$(1/2{c^2}+1/2{\sqrt{b^2+c^2}\sqrt{a^2+c^2}})(-1/2{c^2}+1/2{\sqrt{b^2+c^2}\sqrt{a^2+c^2}})$
To make this easier we separate the negative sign in the second bracket:
(26):$-(1/2{c^2}+1/2{\sqrt{b^2+c^2}\sqrt{a^2+c^2}})(1/2{c^2}-1/2{\sqrt{b^2+c^2}\sqrt{a^2+c^2}})$
Since the brackets are similar to the {(a+b)(a-b)=a^2-b^2} identity we get:
(27):$-(1/4{c^4}-1/4{(b^2+c^2)(a^2+c^2)})$
Multiply out the brackets:
(28):$-(1/4{c^4}-1/4{(a^2)(b^2)+(b^2)(c^2)+(a^2)(c^2)+c^4}$
Multiply by $1/4$:
(29):$-(1/4{c^4}-1/4{(a^2)(b^2)}-1/4{(b^2)(c^2)}-{1/4(a^2)(c^2)}-1/4(c^4))$
Subtract and cancel the negative sign:
(30):$-(-1/4{(a^2)(b^2)}-1/4{(b^2)(c^2)}-1/4{(a^2)(c^2)}$
(31):$1/4{(a^2)(b^2)}+1/4{(b^2)(c^2)}-1/4{(a^2)(c^2)}$
Now we must add the square root sign to the entire equation:
(32):Actual area is $\sqrt{1/4{(a^2)(b^2)}+1/4{(b^2)(c^2)}+1/4{(a^2)(c^2)}}$
(7): The expected area is ${A_4}=\sqrt{1/4{(ab)}^2+1/4{(bc)}^2+1/4{(ac)}^2}$
If you take (32), the actual area, and (7), the expected area, they are the same thing, and thus the theorem has been proved.
Conclusion
The theorem has been proved correct, but not only is this true, this can also bring possibility to "surface area ratios", and even surface area trigonometry. I will try to look more into this in detail.