What determines the height of a tetrahedron?

Question

Tetrahedron $ABCD$ has $AB=AC=AD=BD=17$, $BC=8$, and $CD=15$. Find the volume of $ABCD$.

What are the requirements to determine the height of a $3$D figure if we want to show we have a dihedral angle of $90^{\circ}$. More specifically, say we are given two planes that intersect at a dihedral angle. Is it enough to say that in a tetrahedron as shown below that since the altitude of $ABD$, $AT$, is perpendicular to $DB$ and the median of triangle, $CT$, is perpendicular to $AT$, that $AT$ must in fact be the altitude of the tetrahedron? I am wondering if there is a theorem about this in general. Also, take note that $\triangle{CAT}$ is a right triangle.

If my reasoning above is correct then the answer on the board follows.

Answer 1

What I understand here is that you are making the following argument:

Let $AT$ be an altitude of $ABD$. Since $ABD$ is isosceles, $AT$ is also a median of $ABD$, so $T$ is the midpoint of $BD$. $BCD$ is a right triangle by Pythagoras, hence its median $CT$ has length $\frac{1}{2} BD$. Then you calculate $AT$ and use Pythagoras to prove that $CAT$ has a right angle at $T$. Now you would like to conclude that $AT$ is perpendicular to plane $BCD$.

Yes, you can say this because $AT$ is perpendicular to two intersecting lines in plane $BCD$, namely $CT$ and $BD$. Therefore $AT$ is perpendicular to the whole plane. I'm not sure what the name of this theorem is in English. In Russian it's called the theorem on the three perpendiculars. (The Wikipedia article doesn't link to any other language versions.) In your particular situation, the lines both meet $AT$, but in general this isn't required - both lines could even be skew to $AT$. The important thing is that they're both in plane $BCD$ and not parallel.

However, let me suggest a different approach. If you call $H$ the projection of $A$ onto plane $BCD$, then $AB^2 = AH^2 + HB^2$, $AC^2 = AH^2 + HC^2$, $AD^2 = AH^2 + HD^2$ by Pythagoras. Since $A$ is equidistant from $B$, $C$ and $D$, this shows that $H$ too must be equidistant from $B$, $C$ and $D$. Therefore $H$ must be the centre of the circumscribed circle to $BCD$. In the case of a right triangle, this is the midpoint of the hypotenuse.

I didn't understand what you were saying about a dihedral angle of $90^{\circ}$, or how that's relevant to this problem. A dihedral angle is between two planes (or two half-planes).