ABCD is a trapezoid with ∠DAB=∠ABC=π/2, AD=1 and PA=AB=BC=2; PA is perpendicular to plane ABCD. M is the midpoint of PC. Prove CB is perpendicular to plane PAB.

01 / Standard#Your turn

Official paper · jm02-2021 · 1(b)(i) · PDF 3

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

First prove AD is perpendicular to plane PAB.

Hint 2

AD is parallel to BC in the trapezoid.

Worked solution

AD lies in the base plane, so the perpendicular PA is perpendicular to AD.
$AD\perp PA$
The given angle gives a second intersecting line in plane PAB.
$AD\perp AB,\quad PA\cap AB=\{A\}$
A line perpendicular to both intersecting lines is perpendicular to their plane; transfer this direction to BC.
$AD\perp\operatorname{plane}PAB,\quad BC\parallel AD\implies CB\perp\operatorname{plane}PAB$

CB⊥plane PAB.

Checks and common pitfalls: Perpendicularity to only AB does not prove perpendicularity to the plane.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
AD lies in the base plane, so the perpendicular PA is perpendicular to AD.
$AD\perp PA$
The given angle gives a second intersecting line in plane PAB.
$AD\perp AB,\quad PA\cap AB=\{A\}$
A line perpendicular to both intersecting lines is perpendicular to their plane; transfer this direction to BC.
$AD\perp\operatorname{plane}PAB,\quad BC\parallel AD\implies CB\perp\operatorname{plane}PAB$

Think first. Reveal a hint when the class is ready.

2021 JM02