Solid Geometry – Problem 6: Prove that plane

In the quadrilateral pyramid $P-ABCD$, $AB\parallel CD$, $AB\perp AD$, $BC=CD=2$, $AB=1$, and $\triangle PAD$ is equilateral. Points $M,N$ are the midpoints of $BC,PD$, respectively.

(I) Prove that $MN\parallel$ plane $PAB$.

(II) If the dihedral angle $P-AD-C$ is $\dfrac\pi3$, find the tangent of the angle between line $MN$ and plane $PAD$.

(I) Let $E$ be the midpoint of $AD$, and connect $EN,EM$. By midpoint relations, $EN\parallel AP$, $EM\parallel AB$. Since $AP\cap AB=A$, we have plane $EMN\parallel$ plane $PAB$. Also $MN\subset$ plane $EMN$, so $$MN\parallel\text{plane }PAB.$$

(II) Connect $PM$. From the problem condition, $\angle PEM$ is the plane angle of the dihedral angle $P-AD-C$, so $\angle PEM=\dfrac\pi3$. In $\triangle PEM$, this gives an equilateral triangle with side length $\dfrac{\sqrt3}{2}$. Through $M$, draw $MF\perp PE$ at $F$. Then $$MF=\frac{3\sqrt3}{4}.$$ Also $MF\perp$ plane $PAD$. Since $N,F$ are midpoints of $PD,PE$, $$NF=\frac12DE=\frac{\sqrt3}{4}.$$ $\angle MNF$ is the angle between line $MN$ and plane $PAD$, therefore $$\tan\angle(MN,\,PAD)=\frac{MF}{NF}=3.$$

(I) Using midpoint-line parallel relations, we show plane $EMN\parallel PAB$, and hence $$MN\parallel\text{plane }PAB.$$

(II) From the dihedral-angle section and right-triangle lengths, we get $MF=\frac{3\sqrt3}{4}$, $NF=\frac{\sqrt3}{4}$. Thus $$\tan\angle(MN,\,PAD)=\frac{MF}{NF}=3.$$