Solid Geometry Advanced – Problem 2: Prove that plane

In $\triangle MBC$, $BM\perp BC$. Points $A$ and $D$ are the midpoints of $MB$ and $MC$ respectively, and $BC=AM=2$. Fold $\triangle MAD$ along line $AD$ to a new position $\triangle PAD$ in space, so that $P$ is the image of $M$ and $PA\perp AB$.

Prove that $PA\perp$ plane $ABCD$.

(1) Before folding, points $A,B,C,D$ all lie in the original plane of $\triangle MBC$, so plane $ABCD$ is that base plane.

(2) Since $A$ and $D$ are midpoints of $MB$ and $MC$, segment $AD$ is the midline in $\triangle MBC$. Hence $$AD\parallel BC.$$ (3) Given $BM\perp BC$, we also have $BM\perp AD$. Because $A\in MB$, the segment $MA$ is collinear with $BM$, so $$MA\perp AD.$$ (4) Folding $\triangle MAD$ around the axis $AD$ is a rigid rotation that keeps line $AD$ fixed. Therefore the angle between $MA$ and $AD$ is preserved, and the image segment $PA$ satisfies $$PA\perp AD.$$ (5) The condition in the problem also gives $PA\perp AB$. Since $AB$ and $AD$ are two intersecting lines in plane $ABCD$, a line perpendicular to both is perpendicular to the plane. Thus $$PA\perp \text{plane }ABCD.$$

Because $PA\perp AB$ and $PA\perp AD$ with $AB,AD\subset$ plane $ABCD$, it follows that $PA\perp$ plane $ABCD$.