E–ABCD has rhombus base AB=2√2 and ∠DAB=π/3. EB=EC, ∠BEC=π/2, DE=√5. F is the midpoint of BC; G is the perpendicular foot from E to DF. Prove EG is perpendicular to plane ABCD.

01 / Standard#Your turn

Official paper · jm02-2023 · 1(b) · PDF 3

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Find FG, EG and BG using the right triangles.

Hint 2

Use the converse of Pythagoras in EGB.

Worked solution

From part (a), compute the squared lengths.
$FG=EF\cos\angle DFE=\sqrt6/4,\quad EG^2=2-6/16=13/8$
DF⊥BC, so BFG is right at F; EB²=4.
$BG^2=BF^2+FG^2=2+6/16=19/8,\quad EG^2+BG^2=4=EB^2$
Thus EG⊥GB. It is also perpendicular to DF by construction, and GB,DF are intersecting base-plane lines; hence EG⊥plane ABCD.

EG⊥plane ABCD.

Checks and common pitfalls: Verify two intersecting lines in the base plane, not only DF.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
From part (a), compute the squared lengths.
$FG=EF\cos\angle DFE=\sqrt6/4,\quad EG^2=2-6/16=13/8$
DF⊥BC, so BFG is right at F; EB²=4.
$BG^2=BF^2+FG^2=2+6/16=19/8,\quad EG^2+BG^2=4=EB^2$
Thus EG⊥GB. It is also perpendicular to DF by construction, and GB,DF are intersecting base-plane lines; hence EG⊥plane ABCD.

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

2023 JM02