Solid Geometry – Problem 12: Prove that

In tetrahedron $ABCD$, it is given that $AB\perp$ plane $BCD$. Let $F$ be the midpoint of $CD$. Define planes $$\alpha=\text{plane }ABF,\qquad \beta=\text{plane }BCD.$$

(1) Prove that $\alpha\perp\beta$.

(2) Find the dihedral angle between $\alpha$ and $\beta$.

Build a 3D rectangular coordinate system with $B$ as the origin. Let plane $BCD$ be the $xy$-plane $(z=0)$, and let line $AB$ be the $z$-axis. Take $$B(0,0,0),\ C(1,0,0),\ D(0,1,0),\ A(0,0,h)\ (h>0).$$ Then plane $\beta$ has equation $z=0$, so a normal vector is $\vec n_\beta=(0,0,1)$. The direction vector of line $AB$ is also $(0,0,1)$, hence $$AB\perp\text{plane }\beta.$$ Since $AB\subset$ plane $\alpha$, plane $\alpha$ contains a line perpendicular to plane $\beta$. Therefore $$\alpha\perp\beta.$$ Hence the dihedral angle between $\alpha$ and $\beta$ is $$90^\circ.$$

$\alpha\perp\beta$, so the dihedral angle between them is $90^\circ$.