STEP 1 OF 23 · Lesson Opening
Today: Similar Triangles — the Basics
Same shape, different size. The single most-reused geometric weapon in AIMO Q3–Q5 — and the gateway to the side-ratio-squared area rule.
📌 What you will learn today
How this lesson is structured
- Phase 0 (Steps 2–3): Prerequisites — proportions and the basic-proportionality (parallel-cut) intuition.
- Phase 1 (Steps 4–5): Visual intuition — the enlargement picture & the parallel-cut picture.
- Phase 1.5 (Step 6): Formula handbook (AA, parallel-cut, $k^2$, shared-angle).
- Phase 2 (Steps 7–9): Three guided derivations of the $k^2$ area rule.
- Phase 3 (Steps 10–14): Five worked examples ⭐⭐ → ⭐⭐⭐⭐⭐.
- Phase 4 (Step 15): Five practice problems (P1–P5) with Hint / Answer / Solution.
- Phase 5 (Steps 16–17): Two real AIMO past papers — full Observe / Strategy.
- Phase 5.5 (Step 18): One synthesis problem combining AA + parallel-cut + $k^2$.
- Step 19: Atomic-skill matrix + 8 micro-validations.
- Steps 20–22: Cheat sheet, ⭐ self-assessment, error book.
- Step 23: Bridge forward to Part 3 (Cevian Theorems).
Pedagogy: Every AIMO problem in Phase 5 is wrapped in the same 5-step Observe template — keywords / known / unknown / intermediate / hidden constraint — so when you meet the next unseen similarity problem you have a checklist instead of a guess.
STEP 2 OF 23 · Phase 0 · Prerequisite
Prerequisite ① — Proportions $\dfrac{a}{b} = \dfrac{c}{d}$
If proportions are not yet a reflex, the rest of today won't stick.
The single rule you need
$\dfrac{a}{b} = \dfrac{c}{d} \;\;\Longleftrightarrow\;\; ad = bc$
Cross-multiplication
Whenever two ratios are equal, the products of the diagonals are equal.
Three more identities you should be able to write down without thinking:
- $\dfrac{a}{b} = \dfrac{c}{d} \Rightarrow \dfrac{a+b}{b} = \dfrac{c+d}{d}$ (componendo)
- $\dfrac{a}{b} = \dfrac{c}{d} \Rightarrow \dfrac{a-b}{b} = \dfrac{c-d}{d}$ (dividendo)
- $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{a+c}{b+d}$ (mediant)
Mini-problem (verify you remember)
If $\dfrac{x}{12} = \dfrac{5}{8}$, find $x$.
If $\dfrac{AD}{AB} = \dfrac{2}{5}$ and $AB = 20$, find $AD$.
🔑 Reading trick. Whenever you see "in the ratio $2 : 3$", silently translate to $\dfrac{2}{5}$ of the whole and $\dfrac{3}{5}$ of the whole. Most ratio mistakes come from confusing "part : part" with "part : whole".
STEP 3 OF 23 · Phase 0 · Prerequisite
Prerequisite ② — A line parallel to one side cuts the other two proportionally
Also called the Basic Proportionality Theorem (Thales). Every parallel-cut argument today rests on this.
Statement. In $\triangle ABC$, let $D$ be on $AB$ and $E$ on $AC$. If $DE \parallel BC$, then
$\dfrac{AD}{DB} \;=\; \dfrac{AE}{EC} \quad\text{and}\quad \dfrac{AD}{AB} \;=\; \dfrac{AE}{AC} \;=\; \dfrac{DE}{BC}.$
The first form ("part : part") is most useful when you know two of the four pieces. The second form ("part : whole") is what you need to multiply across to similar triangles.
Why is it true?
Slide $\triangle ADE$ along the line $DE$ — every horizontal slice between $A$ and $DE$ scales by the same factor $\dfrac{AD}{AB}$, because parallel lines preserve ratios. Formally this is the AA criterion (next step) applied to $\triangle ADE$ and $\triangle ABC$, since they share angle $A$ and the parallel lines force matching corresponding angles at $D$ and $E$.
In $\triangle ABC$ with $DE \parallel BC$, $AD = 4$, $DB = 6$, $AE = 6$. Find $EC$.
Same triangle as above. If $BC = 15$, find $DE$.
STEP 4 OF 23 · Phase 1 · Visual Intuition
The enlargement picture — similar triangles are the same shape, scaled
If every angle matches and every side is multiplied by the same factor $k$, you have similar triangles.
Below: a small triangle (left) and the same triangle scaled up by $k = 2$ (right). All three angles are preserved. Every side is doubled. The area, as we will prove in Phase 2, is multiplied by $k^2 = 4$.
SVG 1 — similarity by enlargement
Doubling every side multiplies the area by $2^2 = 4$, taking 6 → 24. The angles are unchanged.
🔑 The litmus test for similarity: "Can I lay one triangle exactly on top of the other after scaling (and possibly flipping)?" If yes, they are similar.
STEP 5 OF 23 · Phase 1 · Visual Intuition
The parallel-cut picture — $DE \parallel BC$ creates a tiny similar copy
A line parallel to one side carves off a smaller triangle that has the same shape.
SVG 2 — parallel-cut similarity
The cut $DE \parallel BC$ at $AD : AB = 2 : 5$ creates $\triangle ADE \sim \triangle ABC$ with side ratio $\tfrac{2}{5}$, so its area is $\bigl(\tfrac{2}{5}\bigr)^2 = \tfrac{4}{25}$ of $\triangle ABC$.
The Parallel-Cut Theorem (one-line form)
$DE \parallel BC \;\;\Longrightarrow\;\; \triangle ADE \sim \triangle ABC$
Why?
$\angle A$ is shared; $\angle ADE = \angle ABC$ because $DE \parallel BC$ (corresponding angles). Two pairs of equal angles → AA → similar.
🔑 Pattern to memorise. Whenever you spot a line drawn parallel to a side of a triangle, write down "similar to the whole, ratio = whichever side I know" — that single line cracks 90 % of parallel-cut problems.
STEP 6 OF 23 · Phase 1.5 · Formula Handbook
Formula handbook — pin this page in your head
Four entries. Memorise.
① The AA criterion
If two angles of one triangle equal two angles of another,
then the triangles are similar.
then the triangles are similar.
Read
"Two equal angles is enough — the third is forced and the sides automatically match in ratio."
② Parallel-cut theorem
$DE \parallel BC \;\Longrightarrow\; \triangle ADE \sim \triangle ABC$
Direct consequence
$\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC}$
③ Side ratio $k$ → area ratio $k^2$ ★ core ★
$\dfrac{[\triangle_1]}{[\triangle_2]} \;=\; \left(\dfrac{\text{side}_1}{\text{side}_2}\right)^{\!2} \;=\; k^2$
Sanity check
Linear scale × linear scale = area scale. Stretch every side by $k$ → area by $k^2$.
④ Shared-angle similarity
If two triangles share an angle and the other two pairs of sides
around that angle are in the same ratio (or there is a parallel cut), they are similar.
around that angle are in the same ratio (or there is a parallel cut), they are similar.
In practice
Look for a vertex used by two sub-triangles. That common vertex donates one matching angle for free.
Variable decoder
k= the linear scale factor between two similar triangles (ratio of corresponding sides)[△]= area of triangle~= "is similar to"AA= Angle-Angle similarity criterion
Quick values to know cold
| Side ratio $k$ | $1$ | $\tfrac{1}{2}$ | $\tfrac{2}{3}$ | $\tfrac{3}{4}$ | $\tfrac{2}{5}$ | $2$ | $3$ | $\tfrac{1}{\sqrt 2}$ |
|---|---|---|---|---|---|---|---|---|
| Area ratio $k^2$ | $1$ | $\tfrac{1}{4}$ | $\tfrac{4}{9}$ | $\tfrac{9}{16}$ | $\tfrac{4}{25}$ | $4$ | $9$ | $\tfrac{1}{2}$ |
⚠ The biggest mistake. Forgetting to square the side ratio. If sides are in ratio $3 : 4$, the area ratio is $9 : 16$ — not $3 : 4$, not $6 : 8$. The factor of $k^2$ is the entire reason the topic exists.
STEP 7 OF 23 · Phase 2 · Derivation 1
Derivation ① — concrete: doubling every side multiplies area by 4
Start with a triangle of base $b = 4$ and height $h = 3$. Its area is
$[\triangle] = \tfrac{1}{2} \cdot b \cdot h = \tfrac{1}{2} \cdot 4 \cdot 3 = 6.$
Now scale every side by $k = 2$. Each linear measurement doubles, so the new base is $b' = 8$ and the new height is $h' = 6$. Area:
$[\triangle'] = \tfrac{1}{2} \cdot 8 \cdot 6 = 24 = 4 \cdot 6.$
That is exactly $k^2 = 2^2 = 4$ times the original.
🎯 Verify: $\dfrac{[\triangle']}{[\triangle]} = \dfrac{24}{6} = 4 = 2^2 = k^2$. ✓
[T2-A3·v1] A triangle has area 12. You scale every side by $k = 3$. New area?
[T2-A3·v2] Two similar triangles have areas 9 and 36. Side ratio (small : large)?
STEP 8 OF 23 · Phase 2 · Derivation 2
Derivation ② — concrete: scaling by 3 multiplies area by 9
Take any triangle of area $A$. Stretch every side by $k = 3$. Both the base and the corresponding height triple, so:
$[\triangle'] = \tfrac{1}{2} \cdot (3b) \cdot (3h) = 9 \cdot \tfrac{1}{2} b h = 9A.$
For a worked instance: a 5-12-13 right triangle has area $\tfrac{1}{2}\cdot 5 \cdot 12 = 30$. Tripling sides → 15-36-39 right triangle, area $\tfrac{1}{2}\cdot 15 \cdot 36 = 270 = 9 \cdot 30$. ✓
🎯 Pattern emerging: $k = 1 \to 1$, $k = 2 \to 4$, $k = 3 \to 9$. The general rule must be $k \to k^2$.
A triangle has area 8. The sides are tripled. New area?
Triangles $T_1$ and $T_2$ are similar. $T_1$ has side 4 and area 6; $T_2$ has side 12. Area of $T_2$?
STEP 9 OF 23 · Phase 2 · Derivation 3
Derivation ③ — abstract: side ratio $k$ → area ratio $k^2$
Two similar triangles $\triangle_1$ and $\triangle_2$ with side ratio $k = \dfrac{\text{side}_1}{\text{side}_2}$. Because corresponding heights scale by the same $k$ (heights are linear measurements):
$[\triangle_1] \;=\; \tfrac{1}{2} \cdot b_1 \cdot h_1$.
$[\triangle_2] \;=\; \tfrac{1}{2} \cdot b_2 \cdot h_2 \;=\; \tfrac{1}{2} \cdot \dfrac{b_1}{k} \cdot \dfrac{h_1}{k}$.
$\dfrac{[\triangle_1]}{[\triangle_2]} \;=\; \dfrac{\tfrac{1}{2} b_1 h_1}{\tfrac{1}{2} \cdot \tfrac{b_1}{k} \cdot \tfrac{h_1}{k}} \;=\; k^2$.
$\boxed{\;\dfrac{[\triangle_1]}{[\triangle_2]} \;=\; k^2\;}$
🎯 Why this generalises: the same argument applies in any dimension. In 2-D you square the linear scale; in 3-D (volumes) you cube it. The "linear → squared" step is the entire idea behind every $k^2$ problem on AIMO.
[T2-A3·v3] Two similar triangles have side ratio $5 : 7$. Larger area is 98. Smaller area?
[T2-A3·v4] Two similar triangles have areas 16 and 49. Side ratio (small : large)?
STEP 10 OF 23 · Phase 3 · Worked Example 1 ⭐⭐
Worked Example 1 — parallel-cut, area ratio
WE 1 · ⭐⭐ · skills T2-A2 + T2-A3
WE1. In $\triangle ABC$, the line $DE$ is parallel to $BC$ with $D$ on $AB$ and $E$ on $AC$, where $AD : AB = 2 : 5$. The area of $\triangle ABC$ is $50$. Find the area of $\triangle ADE$.
Type your answer (single number):
💡 Hints — open as needed
"$DE \parallel BC$" → $\triangle ADE \sim \triangle ABC$. Side ratio is $\tfrac{AD}{AB} = \tfrac{2}{5}$. Area ratio is therefore $k^2 = \tfrac{4}{25}$.
Multiply the area of $\triangle ABC$ by $\tfrac{4}{25}$.
Answer: 8
Solution
$DE \parallel BC \Rightarrow \triangle ADE \sim \triangle ABC$ (parallel-cut).
Side ratio $k = \dfrac{AD}{AB} = \dfrac{2}{5}$.
Area ratio $k^2 = \dfrac{4}{25}$.
$[\triangle ADE] = \dfrac{4}{25} \cdot 50 = 8$.
Reflection
Two skills, applied in order: (i) the parallel cut gives similarity, (ii) the side-to-area square rule converts the side ratio to the area ratio. This pair is the workhorse of every "$DE \parallel BC$" problem.
Tried first?
STEP 11 OF 23 · Phase 3 · Worked Example 2 ⭐⭐⭐
Worked Example 2 — shared-angle similarity, find a missing side
WE 2 · ⭐⭐⭐ · skills T2-A1 + T2-A4
WE2. In $\triangle ABC$, point $D$ lies on $AB$ and point $E$ lies on $AC$ with $\angle ADE = \angle ACB$. Given $AD = 4$, $AE = 6$, and $AB = 9$, find $AC$.
Shared angle $\angle A$ + matching second angle ($\angle ADE = \angle ACB$) → $\triangle ADE \sim \triangle ACB$.
Type your answer (single number):
💡 Hints — open as needed
$\triangle ADE$ and $\triangle ACB$ share angle $A$. The given $\angle ADE = \angle ACB$ gives a second pair of equal angles. By AA, the two triangles are similar — but be careful with the correspondence: $A \leftrightarrow A$, $D \leftrightarrow C$, $E \leftrightarrow B$.
Write the side ratios using corresponding pairs: $\dfrac{AD}{AC} = \dfrac{AE}{AB}$. Plug in the three known values and solve for $AC$.
Answer: $AC = 6$
Solution
$\triangle ADE \sim \triangle ACB$ (AA: $\angle A$ shared, $\angle ADE = \angle ACB$).
Correspondence: $A \to A,\;\; D \to C,\;\; E \to B$.
Therefore $\dfrac{AD}{AC} = \dfrac{AE}{AB}$.
$\dfrac{4}{AC} = \dfrac{6}{9} \Rightarrow AC = \dfrac{4 \cdot 9}{6} = 6$.
Reflection
The trap here is the correspondence — $D$ matches $C$, not $B$. Always read off the angle equality and use it to align vertices, then write the proportion.
Tried first?
STEP 12 OF 23 · Phase 3 · Worked Example 3 ⭐⭐⭐⭐
Worked Example 3 — square cut by diagonal + extra segment
WE 3 · ⭐⭐⭐⭐ · skills T2-A1 + T2-A4 (shared angle)
WE3. $ABCD$ is a square of side $12$. The diagonal $BD$ is drawn. $P$ is the foot of the perpendicular from $A$ onto $BD$. Find $AP$.
$\triangle APB \sim \triangle DAB$ — both contain $\angle B$ and a right angle.
Type your answer (decimal or simplified surd):
💡 Hints — open as needed
$\triangle APB$ and $\triangle DAB$ both have a right angle (the perpendicular at $P$ and the corner $\angle DAB$) and share the angle at $B$. So they are similar by AA — a classic shared-angle / shared-right-angle pair.
Use the similarity to write $\dfrac{AP}{AD} = \dfrac{AB}{BD}$. Compute $BD = 12\sqrt{2}$ (diagonal of side-12 square), then solve for $AP$.
Answer: $AP = 6\sqrt{2}$
Solution
In $\triangle APB$ and $\triangle DAB$: $\angle APB = \angle DAB = 90°$ and $\angle B$ is shared.
So $\triangle APB \sim \triangle DAB$ (AA), with correspondence $A \to D,\; P \to A,\; B \to B$.
$\dfrac{AP}{AD} = \dfrac{AB}{BD}$.
$AD = 12$, $AB = 12$, $BD = 12\sqrt{2}$.
$AP = \dfrac{AD \cdot AB}{BD} = \dfrac{12 \cdot 12}{12\sqrt{2}} = \dfrac{12}{\sqrt{2}} = 6\sqrt{2}$.
Reflection
Whenever an altitude is dropped from a right angle to its hypotenuse, two new triangles appear that are similar to the original — and to each other. This is the entire engine behind the geometric-mean theorem.
Tried first?
STEP 13 OF 23 · Phase 3 · Worked Example 4 ⭐⭐⭐⭐
Worked Example 4 — side ratio 3 : 4, area chase
WE 4 · ⭐⭐⭐⭐ · skill T2-A3 (reverse direction)
WE4. Two similar triangles have corresponding sides in the ratio $3 : 4$. The smaller triangle has area $27$. What is the area of the larger triangle?
Type your answer (single number):
💡 Hints — open as needed
Sides are in ratio $3 : 4$ → area ratio is $3^2 : 4^2 = 9 : 16$.
Set up the proportion $\dfrac{27}{[\text{large}]} = \dfrac{9}{16}$ and solve.
Answer: 48
Solution
Side ratio $3 : 4$ → area ratio $9 : 16$.
$\dfrac{27}{[\text{large}]} = \dfrac{9}{16} \Rightarrow [\text{large}] = \dfrac{27 \cdot 16}{9} = 48$.
Reflection
The mistake to avoid: scaling the area by $\tfrac{4}{3}$ instead of $\tfrac{16}{9}$. Always square the linear ratio before applying it to areas.
Tried first?
STEP 14 OF 23 · Phase 3 · Worked Example 5 ⭐⭐⭐⭐⭐
Worked Example 5 — right triangle with parallel cut
WE 5 · ⭐⭐⭐⭐⭐ · skills T2-A2 + T2-A3 + algebra
WE5. In $\triangle ABC$, $\angle A = 90°$. A line through a point $D$ on $AB$, parallel to $BC$, meets $AC$ at $E$. The area of $\triangle ADE$ is $4$ and the area of the trapezium $DBCE$ is $32$. Find the length $AD$ given that $AB = 9$.
$\triangle ABC$ is split into a small similar copy $\triangle ADE$ and a trapezium.
Type your answer (single number):
💡 Hints — open as needed
$DE \parallel BC$ → $\triangle ADE \sim \triangle ABC$. $[\triangle ABC] = [\triangle ADE] + [\text{trapezium}] = 4 + 32 = 36$. So area ratio $\tfrac{[\triangle ADE]}{[\triangle ABC]} = \tfrac{4}{36} = \tfrac{1}{9}$.
Area ratio $= k^2 = \tfrac{1}{9}$ → side ratio $k = \tfrac{1}{3}$. Therefore $AD = \tfrac{1}{3} \cdot AB = 3$.
Answer: $AD = 3$
Solution
Total area: $[\triangle ABC] = 4 + 32 = 36$.
$DE \parallel BC$ → $\triangle ADE \sim \triangle ABC$.
Area ratio $\dfrac{[\triangle ADE]}{[\triangle ABC]} = \dfrac{4}{36} = \dfrac{1}{9}$.
Side ratio $k = \sqrt{\dfrac{1}{9}} = \dfrac{1}{3}$.
$AD = k \cdot AB = \dfrac{1}{3} \cdot 9 = 3$.
Reflection
Two-stage reasoning: (i) get the area ratio by adding the trapezium back to the small triangle, (ii) take the square root to recover the linear ratio. The "add trapezium back" step is the most-missed move on $k^2$ problems.
Tried first?
STEP 15 OF 23 · Phase 4 · Practice (5 problems)
Phase 4 — practice problems
Mix of the four atomic skills (AA, parallel-cut, $k^2$, shared-angle). Try each cold first; open Hint, Answer, Solution as needed.
P1 · ⭐⭐ · T2-A2
P1. In $\triangle ABC$, $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$. If $AD = 3$, $DB = 6$ and $AE = 4$, find $EC$.
Parallel-cut → $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
$EC = 8$
$\dfrac{3}{6} = \dfrac{4}{EC} \Rightarrow EC = 8$.
P2 · ⭐⭐ · T2-A3
P2. Two similar triangles have side ratio $2 : 5$. The smaller has area $20$. Find the area of the larger.
Area ratio is $4 : 25$.
$125$
$\dfrac{20}{[\text{large}]} = \dfrac{4}{25} \Rightarrow [\text{large}] = 125$.
P3 · ⭐⭐⭐ · T2-A1
P3. In $\triangle ABC$ and $\triangle PQR$, $\angle A = \angle P$ and $\angle B = \angle Q$. Given $AB = 6$, $BC = 8$, $PQ = 9$, find $QR$.
AA → $\triangle ABC \sim \triangle PQR$ with $A \to P$, $B \to Q$, $C \to R$. Sides $AB \leftrightarrow PQ$, $BC \leftrightarrow QR$.
$QR = 12$
$\dfrac{BC}{QR} = \dfrac{AB}{PQ} \Rightarrow \dfrac{8}{QR} = \dfrac{6}{9} \Rightarrow QR = 12$.
P4 · ⭐⭐⭐ · T2-A4
P4. In $\triangle ABC$, $D$ lies on $BC$ such that $\angle BAD = \angle BCA$. Show that $\triangle BAD \sim \triangle BCA$, and given $BD = 4$ and $BC = 9$, find $BA$.
Both triangles share angle $B$. Combined with $\angle BAD = \angle BCA$, AA gives similarity. Correspondence: $B \to B$, $A \to C$, $D \to A$, so $\dfrac{BA}{BC} = \dfrac{BD}{BA}$, i.e. $BA^2 = BD \cdot BC$.
$BA = 6$
$BA^2 = BD \cdot BC = 4 \cdot 9 = 36 \Rightarrow BA = 6$. (This is the geometric-mean / "power of a point" identity.)
P5 · ⭐⭐⭐⭐ · T2-A2 + T2-A3
P5. In $\triangle ABC$, $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$. The areas of $\triangle ADE$ and trapezium $DBCE$ are in the ratio $1 : 3$. Find $\dfrac{AD}{AB}$.
$[\triangle ADE] : [\triangle ABC] = 1 : 4$. Take the square root.
$\dfrac{AD}{AB} = \dfrac{1}{2}$
If $[\triangle ADE] = 1$ and $[\text{trapezium}] = 3$, then $[\triangle ABC] = 4$. Area ratio $\tfrac{1}{4} = k^2 \Rightarrow k = \tfrac{1}{2}$, so $\dfrac{AD}{AB} = \dfrac{1}{2}$.
STEP 16 OF 23 · Phase 5 · AIMO Past Paper
AIMO 2003 · Q4 — extending every side by a copy of itself
Exam-format attempt. Try first, then open hints.
AIMO 2003 · Q4 · [3 marks]
$ABC$ is a triangle. $P$ is the point on the extension of $AB$ beyond $B$ such that $BP = AB$. $Q$ is the point on the extension of $BC$ beyond $C$ such that $CQ = BC$. $R$ is the point on the extension of $CA$ beyond $A$ such that $AR = CA$. The area of $\triangle ABC$ is $51$. Find the area of $\triangle PQR$.
Inner $\triangle ABC$ (teal); outer $\triangle PQR$ (yellow). Each side of $ABC$ has been doubled outward.
Your answer:
💡 Need help? Open these in order, only as needed:
(1) Observe — what to notice (5-step model)
1. 关键词识别 (Keywords): "extension by one more copy of itself" → $BP = AB$, $CQ = BC$, $AR = CA$ — every side doubled outward
2. 已知量 (Known): $[\triangle ABC] = 51$. The three "flap" triangles $\triangle BPC$, $\triangle CQA$, $\triangle ARB$ each share one full side with $\triangle ABC$ and have a vertex outside.
3. 未知量 (Unknown): $[\triangle PQR]$
4. 中间量 (Intermediate): The area of each "flap" triangle compared to $[\triangle ABC]$
5. 隐藏约束 (Hidden): $\triangle PQR$ decomposes into $\triangle ABC$ plus the three flaps. So $[\triangle PQR] = [\triangle ABC] + 3 \cdot (\text{flap area})$, and each flap has area $2 \cdot [\triangle ABC]$.
2. 已知量 (Known): $[\triangle ABC] = 51$. The three "flap" triangles $\triangle BPC$, $\triangle CQA$, $\triangle ARB$ each share one full side with $\triangle ABC$ and have a vertex outside.
3. 未知量 (Unknown): $[\triangle PQR]$
4. 中间量 (Intermediate): The area of each "flap" triangle compared to $[\triangle ABC]$
5. 隐藏约束 (Hidden): $\triangle PQR$ decomposes into $\triangle ABC$ plus the three flaps. So $[\triangle PQR] = [\triangle ABC] + 3 \cdot (\text{flap area})$, and each flap has area $2 \cdot [\triangle ABC]$.
(2) Strategy — how to think about it
Decompose $\triangle PQR$ into the central $\triangle ABC$ plus three outer "flap" triangles. Compute each flap area by the "same height, double base" trick: e.g. $\triangle ABP$ shares its altitude from $C$ with $\triangle ABC$, but has base $AP = 2 \cdot AB$, so $[\triangle BPC] = 2 \cdot [\triangle ABC]$ — wait, more carefully: the flap is $\triangle BPC$ which has base $BP = AB$ and shares the perpendicular from $C$ with $\triangle ABC$, so its area equals $[\triangle ABC]$. Then add the contributions from the cousins, and you find each flap contributes $2 \cdot [\triangle ABC]$. Total: $[\triangle PQR] = 7 \cdot [\triangle ABC]$.
Why this technique generalises: any triangle whose vertices come from "ratio $k$ extensions" of another triangle scales linearly by a fixed factor depending on $k$ — here that factor is exactly $7$.
Why this technique generalises: any triangle whose vertices come from "ratio $k$ extensions" of another triangle scales linearly by a fixed factor depending on $k$ — here that factor is exactly $7$.
(3) Step 1 — area of one flap, e.g. $\triangle BPC$
$\triangle BPC$ has base $BP = AB$ (given). It shares the perpendicular distance from $C$ to line $AB$ with $\triangle ABC$. So $[\triangle BPC] = [\triangle ABC]$.
But the actual flap glued to $\triangle ABC$ on side $BC$ is $\triangle BPQ \cup \triangle BCQ$? Let us be careful and use the cleaner two-stage decomposition (next step).
But the actual flap glued to $\triangle ABC$ on side $BC$ is $\triangle BPQ \cup \triangle BCQ$? Let us be careful and use the cleaner two-stage decomposition (next step).
(4) Step 2 — clean decomposition of $\triangle PQR$
Split $\triangle PQR$ into four pieces by drawing $AB$, $BC$, $CA$:
- Central $\triangle ABC$, area $= 51$.
- $\triangle ARP$ — base $AR = CA$, between $A$, $R$, $P$. Use the lemma: a triangle whose vertices are obtained from $\triangle ABC$ by extending one side by a copy and another by a copy has area $2 \cdot [\triangle ABC]$.
- $\triangle BPQ$ — same lemma, area $= 2 \cdot [\triangle ABC]$.
- $\triangle CQR$ — same lemma, area $= 2 \cdot [\triangle ABC]$.
(5) Step 3 — verify the lemma "flap = $2 \cdot [\triangle ABC]$"
Take the flap $\triangle BPQ$. Vertex $B$, then $P = B + (B-A)$, then $Q = C + (C-B)$. Use the determinant area formula on these three points relative to $\triangle ABC$: with $\vec{u} = B - A$, $\vec{v} = C - A$, $[\triangle ABC] = \tfrac{1}{2}|\vec{u} \times \vec{v}|$. Then $\overrightarrow{BP} = \vec{u}$ and $\overrightarrow{BQ} = -\vec{u} + 2\vec{v} - \vec{u}+\vec{u} \ldots$ — direct expansion gives $|\overrightarrow{BP} \times \overrightarrow{BQ}| = 4 \cdot [\triangle ABC]$, so $[\triangle BPQ] = 2 \cdot [\triangle ABC]$. ✓
Strategy
Decompose $\triangle PQR$ into the central $\triangle ABC$ plus three "flap" triangles. By the same-height-shared-base trick, each flap has area $2 \cdot [\triangle ABC]$, giving the famous "ratio 7" result.
Solution
Place $A, B, C$ at convenient position vectors. By construction $P = 2B - A$, $Q = 2C - B$, $R = 2A - C$.
$\triangle PQR$ decomposes as central $\triangle ABC$ + three corner flaps $\triangle ARP$, $\triangle BPQ$, $\triangle CQR$.
For flap $\triangle BPQ$: $\overrightarrow{BP} = B - A$, $\overrightarrow{BQ} = 2C - 2B = 2(C - B)$.
$|\overrightarrow{BP} \times \overrightarrow{BQ}| = |(B-A) \times 2(C-B)| = 2|(B-A) \times (C-B)| = 2 \cdot 2|(B-A)\times(C-A)| / 2 = 4 [\triangle ABC]$.
So $[\triangle BPQ] = \tfrac{1}{2} \cdot 4 [\triangle ABC] = 2 [\triangle ABC]$. By symmetry the other two flaps are also $2 [\triangle ABC]$ each.
$[\triangle PQR] = [\triangle ABC] + 3 \cdot 2[\triangle ABC] = 7 \cdot 51 = 357$.
$[\triangle PQR] = 357$
Why this technique generalises: the decomposition argument with a fixed ratio $k$ extension on every side always gives $[\triangle PQR] = (3k^2 + 3k + 1) \cdot [\triangle ABC]$. With $k = 1$: $3 + 3 + 1 = 7$. (You can quote this once, then verify for the specific $k$ in the exam.)
💡 Connecting back: This is the side-ratio-to-area-ratio engine (skill T2-A3) applied to three flap triangles simultaneously, then summed. Memorise the result "ratio 1 → factor 7" — it appears on AIMO and AMC almost every decade.
STEP 17 OF 23 · Phase 5 · AIMO Past Paper
AIMO 2019 · Q3 — square diagonal meets a cevian
AIMO 2019 · Q3 · [3 marks]
$ABCD$ is a square of side $24$. The point $P$ lies on side $AB$ with $AP = 8$. The line $DP$ meets the diagonal $AC$ at $Q$. Find the area of $\triangle CQD$.
Shaded yellow: $\triangle CQD$. Side $24$, $AP = 8$, $Q = AC \cap DP$.
Your answer:
💡 Need help? Open these in order, only as needed:
(1) Observe — what to notice (5-step model)
1. 关键词识别 (Keywords): square, diagonal, line $DP$ meets diagonal at $Q$ → look for similar triangles formed by the intersection
2. 已知量 (Known): side $= 24$, $AP = 8$ (so $PB = 16$), $D$ is the corner opposite $B$.
3. 未知量 (Unknown): $[\triangle CQD]$
4. 中间量 (Intermediate): the position of $Q$ along $AC$ (or equivalently along $DP$), the height of $\triangle CQD$ measured from $Q$ to $CD$.
5. 隐藏约束 (Hidden): $\triangle APQ \sim \triangle CDQ$ (vertical angles at $Q$ + alternate angles using $AB \parallel CD$). The ratio is $AP : CD = 8 : 24 = 1 : 3$.
2. 已知量 (Known): side $= 24$, $AP = 8$ (so $PB = 16$), $D$ is the corner opposite $B$.
3. 未知量 (Unknown): $[\triangle CQD]$
4. 中间量 (Intermediate): the position of $Q$ along $AC$ (or equivalently along $DP$), the height of $\triangle CQD$ measured from $Q$ to $CD$.
5. 隐藏约束 (Hidden): $\triangle APQ \sim \triangle CDQ$ (vertical angles at $Q$ + alternate angles using $AB \parallel CD$). The ratio is $AP : CD = 8 : 24 = 1 : 3$.
(2) Strategy — how to think about it
Use the parallel sides $AB \parallel CD$ to identify $\triangle APQ \sim \triangle CDQ$ with ratio $1 : 3$. Then $Q$ divides both $DP$ and $CA$ in ratio $1 : 3$ from the small side. From there, use base $CD = 24$ and the perpendicular height from $Q$ to $CD$ to compute $[\triangle CQD]$.
Why this technique generalises: any time two parallel lines are cut by transversals through a common point, similar triangles with the parallel sides as the matching pair appear automatically.
Why this technique generalises: any time two parallel lines are cut by transversals through a common point, similar triangles with the parallel sides as the matching pair appear automatically.
(3) Step 1 — set coordinates
Place $A = (0, 0)$, $B = (24, 0)$, $C = (24, 24)$, $D = (0, 24)$. Then $P = (8, 0)$.
Line $AC$: $y = x$. Line $DP$: from $(0, 24)$ to $(8, 0)$ has slope $\dfrac{0 - 24}{8 - 0} = -3$, equation $y = 24 - 3x$.
Line $AC$: $y = x$. Line $DP$: from $(0, 24)$ to $(8, 0)$ has slope $\dfrac{0 - 24}{8 - 0} = -3$, equation $y = 24 - 3x$.
(4) Step 2 — find $Q$
Solve $x = 24 - 3x \Rightarrow 4x = 24 \Rightarrow x = 6$, $y = 6$. So $Q = (6, 6)$.
(5) Step 3 — area of $\triangle CQD$
Vertices: $C = (24, 24)$, $Q = (6, 6)$, $D = (0, 24)$.
Base $CD$ is horizontal with length $24$ at height $y = 24$. The perpendicular height from $Q = (6, 6)$ to this line is $24 - 6 = 18$.
$[\triangle CQD] = \tfrac{1}{2} \cdot 24 \cdot 18 = 216$.
Base $CD$ is horizontal with length $24$ at height $y = 24$. The perpendicular height from $Q = (6, 6)$ to this line is $24 - 6 = 18$.
$[\triangle CQD] = \tfrac{1}{2} \cdot 24 \cdot 18 = 216$.
Strategy
Two paths give the same answer: a coordinate sweep, or pure similarity. Both rest on the fact that $AB \parallel CD$ creates $\triangle APQ \sim \triangle CDQ$ at the intersection $Q$.
Solution (coordinates)
Set $A = (0,0)$, $B = (24,0)$, $C = (24,24)$, $D = (0,24)$, $P = (8,0)$.
Diagonal $AC$: $y = x$. Cevian $DP$: $y = 24 - 3x$ (slope $-3$ through $(0,24)$).
Intersection: $x = 24 - 3x \Rightarrow x = 6$, so $Q = (6, 6)$.
Base $CD$ = 24 at height $y=24$; perpendicular height from $Q$ = $24 - 6 = 18$.
$[\triangle CQD] = \tfrac{1}{2} \cdot 24 \cdot 18 = 216$.
Solution (pure similarity, no coordinates)
$AB \parallel CD$, so $\triangle APQ \sim \triangle CDQ$ (vertical angles at $Q$ + alternate interior angles).
Side ratio $\dfrac{AP}{CD} = \dfrac{8}{24} = \dfrac{1}{3}$, so $Q$ divides $DP$ in ratio $1 : 3$ counted from $P$.
Therefore $\dfrac{QD}{DP} = \dfrac{3}{4}$ — but we want the area of $\triangle CQD$ directly.
Triangles $\triangle CDP$ and $\triangle CQD$ share base $CD$. Their heights are proportional to the distances of $P$ and $Q$ from line $CD$, in ratio $4 : 3$.
$[\triangle CDP] = \tfrac{1}{2} \cdot 24 \cdot 24 = 288$ (base $CD$, height $= 24$).
$[\triangle CQD] = \dfrac{3}{4} \cdot [\triangle CDP] = \dfrac{3}{4} \cdot 288 = 216$.
$[\triangle CQD] = 216$
Why this technique generalises: any cevian crossing a diagonal in a parallelogram creates a pair of similar triangles whose ratio comes for free from the parallel sides. Whenever you see "two segments crossing inside a parallelogram", reach for $AB \parallel CD$ first.
💡 Connecting back: This problem fuses skill T2-A4 (shared-angle / vertical-angle similarity) with the parallel-line theorem T2-A2. The coordinate path is the safety net; the similarity path is the elegant exam-time solution.
STEP 18 OF 23 · Phase 5.5 · Synthesis Problem
Phase 5.5 — synthesis problem
Combines the parallel-cut theorem with the side-to-area square rule.
Synthesis · ⭐⭐⭐ · skills T2-A1 + T2-A2 + T2-A3
In $\triangle ABC$, the line $DE$ is drawn parallel to $BC$ with $D$ on $AB$ and $E$ on $AC$, where $AD : DB = 2 : 3$. Find the ratio $[\triangle ADE] : [\triangle ABC]$.
Type the area ratio (e.g.
4:25):💡 Hints — switch toolboxes here
Skill 1 (T2-A2): $DE \parallel BC \Rightarrow \triangle ADE \sim \triangle ABC$.
Skill 2 (T2-A1): The side ratio $k$ is $\dfrac{AD}{AB}$, not $\dfrac{AD}{DB}$.
Skill 3 (T2-A3): The area ratio is $k^2$.
Skill 2 (T2-A1): The side ratio $k$ is $\dfrac{AD}{AB}$, not $\dfrac{AD}{DB}$.
Skill 3 (T2-A3): The area ratio is $k^2$.
(i) Convert $AD : DB = 2 : 3$ into $AD : AB = 2 : 5$. (ii) Square: $4 : 25$.
Solution
[Skill 1 — parallel-cut similarity] $DE \parallel BC \Rightarrow \triangle ADE \sim \triangle ABC$.
[Skill 2 — convert part : part to part : whole] $AD : DB = 2 : 3 \Rightarrow AD : AB = 2 : (2+3) = 2 : 5$.
[Skill 3 — square the side ratio]
$\dfrac{[\triangle ADE]}{[\triangle ABC]} = \left(\dfrac{AD}{AB}\right)^{\!2} = \left(\dfrac{2}{5}\right)^{\!2} = \dfrac{4}{25}$.
$[\triangle ADE] : [\triangle ABC] = 4 : 25$
Lesson: The very common slip is to use $AD : DB$ as the side ratio. The side ratio is always part : whole — convert first, then square.
Tried first?
STEP 19 OF 23 · Atomic Skills + Micro-validation
Atomic skill matrix & quick checks
Four atomic skills, each tested twice. Aim for 8/8 before moving on.
| ID | Skill | Verification value |
|---|---|---|
T2-A1 | AA criterion: two equal angles → similar | Identify and write the correspondence correctly |
T2-A2 | Parallel-cut theorem: $DE \parallel BC \Rightarrow \triangle ADE \sim \triangle ABC$ | $AD = 4, DB = 6, AE = 6 \Rightarrow EC = 9$ |
T2-A3 | Side ratio $k$ → area ratio $k^2$ | Sides $3:4 \Rightarrow$ areas $9:16$ |
T2-A4 | Shared-angle similarity recognition | Common vertex donates one angle for free |
Micro-validation — 2 per skill (8 total)
T2-A1 · AA correspondence
[v1] $\triangle ABC \sim \triangle PQR$ with $A \leftrightarrow P$, $B \leftrightarrow Q$. Given $AB = 5$, $BC = 7$, $PQ = 10$. Find $QR$.
[v2] In $\triangle ABC$ and $\triangle DEF$, $\angle A = \angle D$ and $\angle B = \angle E$. If $AB = 6$ and $DE = 9$, the scale factor (small : large) is?
T2-A2 · Parallel-cut theorem
[v1] In $\triangle ABC$, $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$. $AD = 5$, $DB = 10$, $AE = 4$. Find $EC$.
[v2] In $\triangle ABC$, $DE \parallel BC$, $AD : AB = 3 : 7$, $BC = 14$. Find $DE$.
T2-A3 · Side ratio → area ratio
[v1] Two similar triangles have side ratio $1 : 4$. The smaller has area $5$. Larger area?
[v2] Two similar triangles have areas $25$ and $144$. Side ratio (small : large)?
T2-A4 · Shared-angle similarity
[v1] In $\triangle ABC$, $D$ on $BC$ with $\angle BAD = \angle BCA$. $AB = 6$, $BD = 4$. Find $BC$.
[v2] Right triangle $ABC$ with right angle at $A$, altitude from $A$ meets $BC$ at $H$. If $BH = 4$ and $HC = 9$, find $AH$.
STEP 20 OF 23 · Summary
Cheat sheet — pin this on your wall
① AA Similarity
Two angles equal → triangles similar
Read off the angle equalities, write down the correspondence vertex-by-vertex, then write the side proportions.
② Parallel-Cut Theorem
$DE \parallel BC \Rightarrow \triangle ADE \sim \triangle ABC$
Direct corollary: $\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC}$ and $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
③ Side ratio $k$ → area ratio $k^2$ ★ core ★
$\dfrac{[\triangle_1]}{[\triangle_2]} = k^2$
Always square the linear ratio. Convert "part : part" (e.g. $AD : DB$) to "part : whole" (e.g. $AD : AB$) before squaring.
⚠ Memorise: $k = \tfrac{1}{2} \to \tfrac{1}{4}$; $k = \tfrac{2}{3} \to \tfrac{4}{9}$; $k = \tfrac{3}{4} \to \tfrac{9}{16}$; $k = \tfrac{2}{5} \to \tfrac{4}{25}$; $k = \tfrac{1}{\sqrt 2} \to \tfrac{1}{2}$.
④ Shared-angle / shared-vertex similarity
A vertex used by two sub-triangles donates one matching angle for free. Look for two cevians from the same vertex, or an altitude inside a right triangle.
⑤ Right-triangle altitude
$AH^2 = BH \cdot HC$ (geometric mean)
Altitude from the right angle to the hypotenuse splits the triangle into two smaller triangles, each similar to the original.
⑥ The "ratio 7" decomposition (AIMO 2003 Q4)
Extend each side outward by one copy → outer area $= 7 \cdot$ inner
For general $k$: outer $= (3k^2 + 3k + 1) \cdot$ inner.
⑦ Parallel sides in a square / parallelogram
$AB \parallel CD$ at any cevian intersection $\Rightarrow$ similar triangles with the parallel sides as the matching pair.
⑧ Coordinate fall-back
If the similarity correspondence isn't obvious in 30 seconds, drop coordinates and compute. Coordinates always work; they just cost a minute or two.
STEP 21 OF 23 · Self-assessment
⭐ Self-assessment
Rate your confidence on each atomic skill. Three stars = "I can teach this back in my own words."
① T2-A1 · I can use the AA criterion and write down the correct vertex correspondence.
② T2-A2 · I can apply the parallel-cut theorem (both "part : part" and "part : whole" forms).
③ T2-A3 · I square the side ratio to get the area ratio — and reverse it (take a square root) when I'm given areas.
④ T2-A4 · I can spot shared-angle similarity (common vertex, vertical angles, parallel sides).
⑤ I have walked both AIMO past papers (2003 Q4 and 2019 Q3) covered today.
⭐ 0 / 15 — click stars
STEP 22 OF 23 · Error Book
📒 Your error book — Week 5, Part 2
Every wrong submission today has been silently logged here. Re-do these two more times before the Sunday Mock.
Logged errors this session
No errors yet — submit some AIMO problems to populate this list.
Errors are persisted in sessionStorage.aimoErrors_W5_P2. Reset by closing the browser tab.
Habit to build: when an item appears here twice, write the problem on a physical card and put it on your desk. The third time you re-attempt it cold, you should solve it under 2 minutes.
STEP 23 OF 23 · Wrap-up
🎉 Part 2 complete — Bridge to Part 3
Part 2 of Week 5 complete. You now hold the four pillars of AIMO Similar Triangles I:
- AA criterion — two equal angles is enough
- Parallel-cut theorem — $DE \parallel BC \Rightarrow$ small similar copy at the apex
- Side ratio $k$ → area ratio $k^2$ — the engine of every area-chase
- Shared-angle similarity — common vertex donates one matching angle for free
📌 What's next: Part 3 — Cevian Theorems
Part 3 introduces cevians (lines from a vertex to the opposite side) — medians, altitudes, and angle bisectors. You'll learn:
- Centroid divides each median in ratio $2 : 1$, and the three medians cut the triangle into 6 equal-area pieces.
- Angle bisector theorem: $\dfrac{BD}{DC} = \dfrac{AB}{AC}$.
- Altitude $= \dfrac{2 \cdot \text{Area}}{\text{base}}$ — used to chase unknown lengths from area.
The atomic similarity skills you mastered today are the prerequisite for every cevian proof — especially the angle bisector theorem, which is built directly on a parallel-cut similarity argument.
Before Sunday Mock:
- Re-attempt every item in your error book until you get it right under 2 minutes cold.
- Re-skim the Cheat Sheet (Step 20). Memorise the squared-ratio table.
- Make sure all four atomic skills are 3-star on your self-assessment.
▶ Next: Part 3 — Cevian Theorems