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 DM is parallel to plane PAB. Use N, the midpoint of PB, and show ADMN is a rectangle.

01 / Standard#Your turn

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

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Apply the midpoint theorem in triangle PBC.

Hint 2

Show AD and NM are parallel, equal and perpendicular to AN.

Worked solution

The midpoint segment has the required direction and length.
$NM\parallel BC\parallel AD,\quad NM=BC/2=1=AD$
One pair of equal parallel opposite sides makes ADMN a parallelogram, and AD⊥AN gives a right angle.
$AN\subset\operatorname{plane}PAB,\quad AD\perp AN$
Thus ADMN is a rectangle and DM is parallel to AN.
$DM\parallel AN$
D is outside plane PAB, while AN lies in it. A line through D parallel to AN therefore has no intersection with the plane.
$DM\parallel\operatorname{plane}PAB$

ADMN is a rectangle; DM∥plane PAB.

Checks and common pitfalls: A line parallel to a line in a plane also needs to be outside the plane to establish line–plane parallelism.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The midpoint segment has the required direction and length.
$NM\parallel BC\parallel AD,\quad NM=BC/2=1=AD$
One pair of equal parallel opposite sides makes ADMN a parallelogram, and AD⊥AN gives a right angle.
$AN\subset\operatorname{plane}PAB,\quad AD\perp AN$
Thus ADMN is a rectangle and DM is parallel to AN.
$DM\parallel AN$
D is outside plane PAB, while AN lies in it. A line through D parallel to AN therefore has no intersection with the plane.
$DM\parallel\operatorname{plane}PAB$

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

2021 JM02