In the unit cube shown in the official diagram, M is on EC and GM⊥EC. Prove triangles CMG, CMB and CMD are congruent.

01 / Standard#Your turn

Official paper · jm02-2026 · 1(a) · PDF 3

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Use three mutually perpendicular cube edges as coordinate axes.

Hint 2

Project G onto the body diagonal EC.

Worked solution

Choose coordinates consistent with the original vertex labels.
$E=(0,0,0),\ C=(1,1,1),\ B=(1,0,1),\ D=(0,1,1),\ G=(1,1,0)$
Write M=(t,t,t) and impose the perpendicularity condition.
$(G-M)\cdot(1,1,1)=2-3t=0\implies M=\left(\frac23,\frac23,\frac23\right)$
All three triangles have the same three side lengths, so SSS applies.
$CG=CB=CD=1,\quad CM=\frac1{\sqrt3},\quad MG=MB=MD=\sqrt{\frac23}$

△CMG≅△CMB≅△CMD by SSS.

Checks and common pitfalls: The diagram is not to scale; equality comes from cube geometry or coordinates.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Choose coordinates consistent with the original vertex labels.
$E=(0,0,0),\ C=(1,1,1),\ B=(1,0,1),\ D=(0,1,1),\ G=(1,1,0)$
Write M=(t,t,t) and impose the perpendicularity condition.
$(G-M)\cdot(1,1,1)=2-3t=0\implies M=\left(\frac23,\frac23,\frac23\right)$
All three triangles have the same three side lengths, so SSS applies.
$CG=CB=CD=1,\quad CM=\frac1{\sqrt3},\quad MG=MB=MD=\sqrt{\frac23}$

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

2026 JM02