Is this a sufficient proof of a math contest problem? Problem:
If a,b,c,d are real, prove that 

$$a^2+b^2=2$$
$$c^2+d^2=2$$
$$ac=bd$$

Is true if and only if

$$a^2+c^2=2$$
$$b^2+d^2=2$$
$$ab=cd$$

My proof is as follows:
Note that each of the second set of equations nearly corresponds to the lengths of the sides of a right triangle. The equations may be equivalently written as:
$$\sqrt{a^2+c^2}=\sqrt{2}$$
$$\sqrt{b^2+d^2}=\sqrt{2}$$
$$\frac{c}{a}=\frac{b}{d}$$
It should also be noted that $Arctan(\frac{c}{a})=Arctan(\frac{b}{d})=\theta$. Represented geometrically, the two triangles represented by these equations would look like this:

Note that the two triangles are similar, and that they share the hypotenuse length $\sqrt{2}$. This means that the triangles are congruent, which means that the sides opposite the angle are as well, and the sides adjacent the angle are also congruent.
So one can surmise that $b=c$ and $a=d$. When substituted into the given equations, you get the resulting equations.
This covers the cases of $|a,b,c,d|>0$. If $a=0$, then we know that $b=\sqrt{2}$. If $b=\sqrt{2}$ and $a=0$, then $d\sqrt{2}=0$, so $d=0$. Because $d=0$, we know that $c=\sqrt{2}$.
By symmetry, the same argument for a=0 can apply to b=0 if a and b switch places. Sorry if that's a bit lazy. The remaining arguments are easier, though tedious to prove, and I'm primarily concerned with the geometric argument.
Please let me know if this is sufficient, and if not, where I went wrong. Thank you!
 A: The difficulty with the geometric argument is that you cannot actually conclude that $b=c$ and $a=d$.  For example, if the second set of equations is true, then you can have $$a = \sqrt{1/2}, c = \sqrt{3/2}, \quad b = -\sqrt{3/2}, d = -\sqrt{1/2},$$ which satisfies the conditions but it is clear that $b \ne c$.  This is because your triangles do not distinguish lengths of opposite sign.
A fully rigorous geometric argument would probably be inelegant due to the need to handle such cases where the variables do not have the same sign, or one or more variables is zero.  So an algebraic approach may be more appropriate.  We would note $(ab)^2 = (cd)^2$ as a result of squaring the third equation; then substituting $a^2 = 2-c^2$ and $d^2 = 2-b^2$, we get $$(2-c^2)b^2 = c^2(2-b^2),$$ or $2b^2 = 2c^2$, or $b^2 = c^2$.  Similarly, we easily conclude $a^2 = d^2$.  Then using these relationships in the first two equations gives $$a^2 + b^2 = 2, \quad c^2 + d^2 = 2.$$  The last desired condition readily follows since given $ab = cd$ and $b^2 = c^2$, we have $$b(cd) = (ab)b = ab^2 = ac^2,$$ and if $c \ne 0$, cancellation gives $bd = ac$ as desired.  If $c = 0$, then $b = 0$ as well, thus $ac = 0 = bd$.  This completes the proof.
