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.
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.
The sides of the three right angles are $\displaystyle a,b,c.$Code:* * * * c * * * a* *b * * * *e * * * * * * * * * * d
The sides of the base are $\displaystyle d,e,f.$
We have: .$\displaystyle \begin{Bmatrix}a^2+b^2 &=& d^2 \\ b^2+c^2 &=& e^2 \\ c^2+a^2 &=& f^2 \end{Bmatrix}$
Suppose the base triangle has a right angle,
. . then we have (for example): .$\displaystyle d^2+e^2 \:=\:f^2$
Substitute: .$\displaystyle (a^2+b^2) + (b^2+c^2) \:=\;c^2+a^2$
And we have: .$\displaystyle 2b^2 \:=\:0 \quad\Rightarrow\quad b \:=\:0$
. . There is no tetrahedron.
Therefore, the base triangle canNOT be right-angled.