Relation between edgelengths in a tetrahedron with two right angles and three equal edges I have got a problem I can't solve myself. I had an attempt, but it's wrong. I was told to draw a grid of this tetrahedron and then it's easier to find a solution (I tried it, but I don't see anything).

There is a tetrahedron (ABCD), where 
  $$\angle{ACB}=\angle{ADB}=90^\circ $$ and $$ AC=CD=DB $$
Prove, that $$ AB<2CD  $$

What I've already done (wrong):
I marked
$$
|AC|=|CD|=|DB|=a
$$
$$
|AB|=c
$$
$$
|CB|=b
$$
I'm proving, that AB<2CD, so it means c<2a
Using Pythagoras' theorem":
$$
c=\sqrt{a^2+b^2}
$$
so
$$
\sqrt{a^2+b^2}<2a
$$
and after transformation:
$$
b<\sqrt3a
$$
In order to prove, that AB<2CD, I need to prove
$$
b<\sqrt3*a
$$
In order to make CDB exist, it has to satisfy the following:
$$
a+b>a
$$
and
$$
a+a>b
$$
so:
$$
2a>b
$$
And now if 
$$b<2a
$$
and
$$
b<\sqrt3*a
$$
then b has to be less than sqrt3*a, because
$$
\sqrt{3}a<2a
$$
I proved, that
$$
b<\sqrt3a
$$
so
$$
c<2a
$$
so
$$
AB<2CD
$$
 A: Let's put the tetrahedron vertices on a 3-D Cartesian coordinate system.  Without loss of generality, we can choose $C$ to be at the origin.  
We can then choose our unit of length in our coordinate system such that $AC = 1$; again, this loses no generality because it is just a matter of choosing our coordinate markings, without changing the tetrahedron.  And we can choose the $X$ direction such that in fact
$A$ is at $(1,0,0)$.
Since $\angle ACB = 90^\circ$, the $X$ coordinate of $B$ is constrained to be zero, but we are free to draw the $Y$ direction in the same direction as $CB$. Since $CD = AC = 1$ this places $D$ at $(0,b,0)$ for some (non-zero) value of $b$.   We can also label the coordinates of $D$ as $(x,y,z)$.
So there is some coordinate system such that the vertices are at
$$
\begin{array}{cccc}
A = (1,0,0) & B = (0,b,0) & C = (0,0,0) & D = (x, y, z)
\end{array}
$$
Then we have three further pieces of information:
$$
\begin{array}{lcl}
CD = AC =1 &\Longrightarrow & x^2 + y^2 +z^2 = 1 \\
DB = AC =1 &\Longrightarrow & x^2 + (y-b)^2 +z^2 = 1 \\
\angle ADB = 90^\circ & \Longrightarrow & (x-1)x + y(y-b) + z^2 = 0
\end{array} 
$$
The first two equations determine 
$$ \begin{array}{c} y = \frac{b}{2} \\ x^2+z^2 = 1 - \frac{b^2}{4}\end{array}$$
and substituting those results in the third equation gives 
$$
\begin{array}{c}
x = 1 - \frac{b^2}{2}\\
z = \frac{b}{2}\sqrt{3 - b^2}\\
D = \left( 1 - \frac{b^2}{2},\frac{b}{2}, \frac{b}{2}\sqrt{3 - b^2}\right)
\end{array}
$$
A key observation is that for this to work, $z$ has to be real and non-zero (if $z=0$ the tetrahedron collapses into the $XY$ plane).  So we need the restriction that 
$$
0 < |b| < \sqrt{3} $$
Then the relevant side lengths of our our tetrahedron, measured using the unit length of our coordinate system,  are
$$
\begin{array}{c}
AB = \sqrt{b^2+1} \\
CD = 1
\end{array}
$$
In particular, since $|b| < \sqrt{3}$,
$$
AB < \sqrt{3+1} = 2 = 2\, CD
$$
which proves the proposition.
A: The fact that $\angle ACB = \angle ADB = 90^\circ$ means that $C$ and $D$ lie on the sphere which has $AB$ as a diameter. Let $O$ be the center of this sphere, i.e., the midpoint of $AB$. Since $A$ and $B$ lie at the opposite points of the sphere, we have
$$
\angle AOC + \angle COD + \angle DOB > 180^\circ.
$$
(It it strictly greater, since $A,B,C,D$ do not lie in the same plane.) But $AC = CD = DB$ means that $\angle AOC = \angle COD = \angle DOB$, so $\angle COD>60^\circ$. 
Now let $r$ be the radius of the sphere, then $AB = 2r$, and $CD = 2r\sin\frac{\angle COD}{2}>2r\sin 30^\circ = r$.
A: Since $\angle ACB=\angle ADB=90^{\circ}$, both $C$ and $D$ lie on a circle (which is not unique) with diameter $AB$. Since the locus of all possible circles with a certain diameter is a sphere (by definition), $A$, $B$, $C$, $D$ form a sphere and $AB$ is a diameter. 
Let $O$ be the center of the circle (which is effectively the midpoint of $AB$. 
Since $AC=CD=DB$, $\angle AOC=\angle DOB=\angle COD$. If $A$, $B$, $C$, $D$ are coplanar, then the sum of $\angle AOC$, $\angle DOB$, and $\angle COD$ is $180^{\circ}$. However,they determine a tetrahedron, so they are not coplanar, so $\angle AOC+\angle DOB+\angle COD>180^{\circ}$. Since their measures are equal, all of them are greater than $60^{\circ}$. 
Hence $\angle COD>60^{\circ}$, and if we consider a circle and we draw both $CO$ And $DO$ (let them be $r$, then $CD^2=2r^2-2r^2\cos{\angle COD}$. Note that $\cos{60}=\frac{1}{2}$, making $CD=r$, with $AB=2CD$. The $\cos$ function is decreasing from $0^{\circ}$ to $180^{\circ}$, so if $\angle COD>60^{\circ}$ then $CD>r$. Since $AB=2r$, $AB<2CD$. 
A: The problem is visualized by folding card-board triangles along line/hypotenuse $AB$, which is the diameter of sphere.There are three connected chords of equal length, we picturized and formulate how the line CD moves.
EDIT
It is seen that the situation is more amenable for handling using spherical trigonometry. Lines Ac, CB and BA are central, other three lines are simply joined once the  latitude/longitude, meridian are followed. So, it is re-cast in the following way:
Let ACDB be the Z shaped zigzag connected path of chords which can be placed inside corresponding great circles. It is convenient to derive CD by taking parts of  meridians through North and South poles and another intersecting great circle so as to satisfy all given conditions.
Spherical coordinates  $ r, \theta, \phi $ ( radius, longitude, latitude of parallel circle ) are used. 
( The same symbol $ \phi $ used to denote starting angle of North hemisphere parallel circle latitude. )
AC runs between North pole $ A (a,0, \pi/2)$ to $C(a, 0, \phi)$,
CD is between $ C ( a,0, \phi)$ to $ D ( a,  \Delta \theta, -\phi)$,and,
DB between $ D ( a,\Delta \theta, -\phi)$ to South pole $B ( a, \Delta \theta, -\pi/2 )$.
AB is the diameter $ 2 a $ from North pole to South pole. 
Angle subtended between two points at center of sphere radius $a$ is given by
$$ \cos \omega = \cos \phi_1\cos \phi_2 \cos ( \Delta \theta) + \sin \phi_1\sin \phi_2 $$
From the aforesaid and the given data, 
$$ \phi_1= \phi,\,\phi_2= -\phi,\,\omega= \pi/2-\phi ..(1*) $$
We have
$$ \cos \Delta \theta = \tan \phi ( \tan \phi   + \sec \phi) ...(2*) $$
$$ \dfrac {CD}{a} = 2 \sin ( \Delta \theta/2) = \sqrt{ 2 ( 1- \cos \Delta \theta  )} $$
$$ \dfrac{CD}{ a} =  \sqrt{ 2 ( 1- \tan \phi\,(\tan \phi+ \sec \phi) \, )}  ...(3*) $$
It is insightful to introduce $ \phi_c = \pi/2 - \phi $, which is referred to as the 
co-latitude from pole instead of equator.
So we can alternately express the relations: 
$$ CD = 2 a \sin (\phi_c/2) , \cos \Delta \theta = \cos \phi_c / ( 1-\cos \phi_c) $$
and so, essentially,
$$ CD/a = \sec (\Delta \theta /2) > 1, always ...(4*) $$   
The OP mentions his possible error. This answer accordingly attempts to comprehensively 
contain all possible cases we get by folding the meridional planes through polar/azimuth 
angle $ \Delta \theta $.

Three special cases  are mentioned( dots on graph), two are illustrated, everything else is covered by connecting equations.
by given equations and graphs given here.
Typical case

The above 3D diagram  has been drawn after numerical calculation from above expression (3) for starting  $ \phi$  = 25 degrees, co-latitude 65 degrees, each chord length 1.072 a. Angle between meridional planes by virtue of (2*) calculates to 42.9 degrees.Each of three chords (AC,CD and DB ) subtend 65 degrees at center of sphere.
Special case 1
When $\Delta \theta$ = 90 degrees, $\phi$ =0, AC goes from North pole to Equator, then to D executing a quarter of equatorial circumference, then down to B, the South pole. Length of each chord is $ \sqrt 2 a $. This tetrahedron is one-fourth of a regular octahedron.
Special case 2
This occurs when $ \Delta \theta =0 , \phi= \pi/6 $ or 30 degrees. ACDB is half of a flat regular hexagon as shown below.

Conclusion : The expression $ CD = a \sec(\Delta \theta) > a $ quantitatively establishes  AB < 2 CD for all possible configurations in the tetrahedron with the given conditions.
A: The volume of a tetrahedron with edge lengths $a$, $b$, $c$ and face angles $\alpha$, $\beta$, $\gamma$ as shown

is given by
$$V = \frac{abc}{6}\sqrt{1+2\cos\alpha\cos\beta\cos\gamma - \cos^2\alpha-\cos^2\beta-\cos^2\gamma}$$
For the situation in question, we have
$$\cos\alpha = \frac{b}{c}\qquad\cos\beta = \frac{a}{c}\qquad\cos\gamma = \frac{b}{2a} \qquad c^2 = a^2 + b^2$$
so that
$$V = \frac{abc}{6}\sqrt{\require{cancel}\cancel{1} + 2 \frac{ab^2}{2ac^2} - \require{cancel}\cancel{\frac{b^2}{c^2}} - \require{cancel}\cancel{\frac{a^2}{c^2}} - \frac{b^2}{4a^2}} = \frac{abc}{6}\sqrt{\frac{b^2}{c^2} - \frac{b^2}{4a^2}} = \frac{b^2}{12}\sqrt{4a^2 - c^2}$$
For a real volume, we must have $4a^2 \geq c^2$, so that (for positive edge-lengths) $2a\geq c$. $\square$
