CAPS EXAM-FOCUSED STUDY NOTES Paper 2 | Euclidean Geometry
Ratio, Proportion and Similarity
CAPS SCOPE
This topic covers the Grade 12 Euclidean Geometry content on ratio, proportion and similarity, as
examined in Paper 2. It builds on the Grade 10 midpoint theorem and Grade 11 circle geometry,
and is usually combined with circle geometry riders in the final Euclidean Geometry question.
You must be able to:
• State and apply the Proportionality Theorem and its converse.
• Prove the Proportionality Theorem (examinable proof — can be asked directly in Paper 2).
• Identify and prove triangles similar using the correct conditions.
• Use similarity to calculate unknown lengths and ratios.
• Apply the ratio of areas of similar triangles.
• Combine ratio/proportion/similarity with circle geometry in a single rider.
Connects to: Grade 10 midpoint theorem, Grade 11 circle geometry (tan-chord, cyclic quads), and the
co-ordinate geometry section of Paper 2 where similarity sometimes supports a proof.
WORKING WITH RATIO, PROPORTION AND SIMILARITY
A ratio such as AD : DB compares two lengths in a fixed order — the order you're given is the order
your answer must keep. When two ratios are set equal to each other, e.g. AD/DB = AE/EC, that
equation is a proportion. Almost every mark in this topic comes from writing that proportion
statement explicitly, not just assuming it.
“Similar” has a specific meaning here: two triangles (or polygons) are similar when they are the
same shape but not necessarily the same size — corresponding angles are equal and
corresponding sides are all in the same ratio. The notation ΔDEF ||| ΔABC encodes the
correspondence directly in the letter order: D matches A, E matches B, F matches C. Writing the
letters in the wrong order is one of the most common ways marks are lost, even when the
geometry reasoning is correct.
THEOREM 1 — THE PROPORTIONALITY THEOREM
Statement
PROPORTIONALITY THEOREM
A line drawn parallel to one side of a triangle divides the other two sides proportionally.
In ΔABC, if DE // BC (D on AB, E on AC), then:
AD/DB = AE/EC
StudyForge — Ratio, Proportion and Similarity (Grade 12 Paper 2) Page 1
, STUDYFORGE Grade 12 Mathematics
CAPS EXAM-FOCUSED STUDY NOTES Paper 2 | Euclidean Geometry
Ratio, Proportion and Similarity
A
D E DE // BC
B C
Proof (examinable — learn this proof)
Given: ΔABC with DE // BC, D on AB and E on AC.
Required to prove: AD/DB = AE/EC
• Construction: Join DC and BE. Draw perpendicular height h1 from E to AB, and perpendicular
height h2 from D to AC.
• Area ΔADE = ½ × AD × h1 and Area ΔBDE = ½ × DB × h1 → same perpendicular height h1, so
AreaΔADE / AreaΔBDE = AD/DB.
• Similarly, Area ΔADE = ½ × AE × h2 and Area ΔCDE = ½ × EC × h2 → AreaΔADE / AreaΔCDE =
AE/EC.
• ΔBDE and ΔCDE stand on the same base DE and lie between the same parallel lines DE and BC,
so they have equal height → AreaΔBDE = AreaΔCDE.
• Therefore AreaΔADE/AreaΔBDE = AreaΔADE/AreaΔCDE, which gives AD/DB = AE/EC. ∎
EXAM TIP If a proof question is worth 5 marks, the marking points usually match the 5 bullet
steps above: construction, two area ratios linked to equal heights, the equal-area statement for
ΔBDE and ΔCDE, and the final conclusion. Write each step as its own line — do not compress
steps.
Converse
CONVERSE OF THE PROPORTIONALITY THEOREM
If a line divides two sides of a triangle proportionally, then that line is parallel to the third side.
In ΔABC, if AD/DB = AE/EC, then DE // BC.
The converse is used constantly in riders to prove parallel lines — whenever a question gives you
two equal ratios on the sides of a triangle and asks you to prove a line is parallel to the third side,
this is the theorem to quote.
QUICK RECALL DE // BC ↔ AD/DB = AE/EC. Also true: AD/AB = AE/AC (whole side to whole
side).
THEOREM 2 — SIMILAR TRIANGLES
StudyForge — Ratio, Proportion and Similarity (Grade 12 Paper 2) Page 2