Proof in a tetrahedron - I added my wrong attempt

Question

please help me solve this problem:

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

Prove, that $$ AB<2CD $$

My (wrong) attempt:

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 $$ AB<2CD $$

Best regards, Adam.

Answer 1

It is impossible, because $AB$ is the hypotenuse of $\bigtriangleup ABC$, then $AB>AC=CD$.

Answer 2

It is correct to state that "In order to prove, that AB < 2CD, I need to prove $b < \sqrt3a$". Your proof is also correct about $b < 2a$. But you cannot prove $b < \sqrt3a$ simply because $b < 2a$.

For me, I can only prove $AB < \sqrt5CD$ as in the following:

      $b < 2a$
=> $b^2 < 4a^2$
=> $a^2+b^2 < 5a^2$
=> $\sqrt{a^2+b^2} < \sqrt{5}a$
=> $c < \sqrt{5}a$.