1. ## Tetrahedron proof

Could I have a clue how to start this?

The edges of a tetrahedron meet at a vertex so that a right angle is formed between each pair of edges. Prove that the base triangle cannot be right angled.

2. ## Re: Tetrahedron proof

Originally Posted by Stuck Man
Could I have a clue how to start this?

The edges of a tetrahedron meet at a vertex so that a right angle is formed between each pair of edges. Prove that the base triangle cannot be right angled.
A tetrahedron has six edges. Not all edges meet at the same vertex. This question does not make sense to me.

3. ## Re: Tetrahedron proof

Hello, Stuck Man!

Did you make a sketch?

Three edges of a tetrahedron meet at a vertex so that a right angle
is formed between each pair of edges.
Prove that the base triangle cannot be right-angled.

Code:
            *
* * * c
*   *   *
a*     *b    *
*       *   *e
*         * *
* * * * * * *
d
The sides of the three right angles are $a,b,c.$
The sides of the base are $d,e,f.$

We have: . $\begin{Bmatrix}a^2+b^2 &=& d^2 \\ b^2+c^2 &=& e^2 \\ c^2+a^2 &=& f^2 \end{Bmatrix}$

Suppose the base triangle has a right angle,
. . then we have (for example): . $d^2+e^2 \:=\:f^2$

Substitute: . $(a^2+b^2) + (b^2+c^2) \:=\;c^2+a^2$

And we have: . $2b^2 \:=\:0 \quad\Rightarrow\quad b \:=\:0$

. . There is no tetrahedron.

Therefore, the base triangle canNOT be right-angled.