To prove two triangles similar you do not need to check three angles and three ratios. Two equal angles (AA), three proportional sides (SSS), or one equal angle between two proportional sides (SAS) are enough.
In Class 9 you learned that two triangles are congruent without checking all six pairs of parts: SSS, SAS or ASA is enough. Similarity works the same way. Definition 6.3 asked for three equal angles and three equal ratios of sides, but a surveyor measuring a tower cannot check six things — and does not need to. A short, reliable test is enough.
In this section you meet three such tests, called the criteria for similarity of triangles: AA (two angles), SSS (three sides in proportion) and SAS (one angle and the two sides around it in proportion). You learn when each applies, which tempting "tests" do not work, and how to use them to find lengths in shadow problems, crossing ladders, overlapping triangles and triangles cut by parallel lines.
The AA criterion
📐 Theorem 6.3 — AA (angle–angle) criterion
If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar. That is, if $\angle A=\angle D$ and $\angle B=\angle E$ in $\triangle ABC$ and $\triangle DEF$, then $\triangle ABC\sim\triangle DEF$.
The third angles are automatically equal too, because each triangle's angles add up to $180^\circ$: $\angle C=180^\circ-\angle A-\angle B=180^\circ-\angle D-\angle E=\angle F$. For this reason the criterion is also called the AAA criterion.
Why it works (outline). Mark $P$ on $AB$ and $Q$ on $AC$ with $AP=DE$ and $AQ=DF$. Since $\angle A=\angle D$, the triangles $APQ$ and $DEF$ are congruent (SAS), so $\angle APQ=\angle E=\angle B$. Equal corresponding angles mean $PQ\parallel BC$, and the Basic Proportionality Theorem (Theorem 6.1) gives $\dfrac{AB}{AP}=\dfrac{AC}{AQ}$, i.e. $\dfrac{AB}{DE}=\dfrac{AC}{DF}$. Repeating the argument from vertex $B$ gives $\dfrac{AB}{DE}=\dfrac{BC}{EF}$. So all three ratios are equal, and the angles are equal: the triangles are similar.
The AA criterion justifies the shadow argument of Section 6.1: a stick and a tower both make a right angle with the ground, and the Sun's rays make the same angle with the ground at both places, so two angles match. It is also the criterion you will use most, because angles are easy to spot: right angles, vertically opposite angles, a shared angle, or angles formed by parallel lines.
✏️ Example 1 — Applying AA
In $\triangle ABC$, $\angle A=50^\circ$ and $\angle B=60^\circ$. In $\triangle PQR$, $\angle P=50^\circ$ and $\angle R=70^\circ$. Are the triangles similar? If so, write the statement.
💡 Show solution▶
Third angles $\angle C=180^\circ-50^\circ-60^\circ=70^\circ$ and $\angle Q=180^\circ-50^\circ-70^\circ=60^\circ$.
Compare $\angle A=\angle P=50^\circ$, $\angle B=\angle Q=60^\circ$ (and $\angle C=\angle R=70^\circ$). Two pairs are equal, so the AA criterion applies.
Statement $\triangle ABC\sim\triangle PQR$ — the vertices are listed in matching order $A\leftrightarrow P$, $B\leftrightarrow Q$, $C\leftrightarrow R$. (Writing $\triangle ABC\sim\triangle PRQ$ would be wrong.)
✏️ Example 2 — Parallel lines give similar triangles
In $\triangle ABC$, $D$ is on $AB$ and $E$ is on $AC$ with $DE\parallel BC$. (a) Show that $\triangle ADE\sim\triangle ABC$. (b) If $AD=4$ cm, $AB=10$ cm, $AC=12$ cm and $BC=15$ cm, find $DE$ and $AE$.
💡 Show solution▶
(a) Angles $\angle A$ is common to both triangles. Since $DE\parallel BC$ with transversal $AB$, $\angle ADE=\angle ABC$ (corresponding angles).
(a) Conclude Two pairs of angles are equal, so $\triangle ADE\sim\triangle ABC$ by AA ($A\leftrightarrow A$, $D\leftrightarrow B$, $E\leftrightarrow C$).
(b) Ratios Corresponding sides: $\dfrac{AD}{AB}=\dfrac{AE}{AC}=\dfrac{DE}{BC}$, so $\dfrac{4}{10}=\dfrac{AE}{12}=\dfrac{DE}{15}=0.4$.
(b) Solve $DE=0.4\times15=6$ cm and $AE=0.4\times12=4.8$ cm.
Note The Basic Proportionality Theorem told us only about $AD:DB$ and $AE:EC$; similarity also gives the third side $DE$ in terms of $BC$.
The SSS and SAS criteria
When angles are not given, we can compare sides. The two criteria below say that three proportional sides, or two proportional sides with the angle between them equal, force the triangles to have the same shape.
📐 Theorem 6.4 — SSS (side–side–side) criterion
If the three sides of one triangle are proportional to the three sides of another, that is, $\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{CA}{FD}$, then the triangles are similar: $\triangle ABC\sim\triangle DEF$ (and so their corresponding angles are equal).
📘 Definition 6.4 — Included angle
In a triangle, the angle formed by two given sides at their common vertex is the included angle of those sides. In $\triangle ABC$ the included angle of $AB$ and $AC$ is $\angle A$.
📐 Theorem 6.5 — SAS (side–angle–side) criterion
If one angle of a triangle is equal to one angle of another triangle, and the two sides including that angle are proportional to the two sides including the equal angle in the other triangle, then the triangles are similar. That is, if $\angle A=\angle D$ and $\dfrac{AB}{DE}=\dfrac{AC}{DF}$, then $\triangle ABC\sim\triangle DEF$.
We state these two criteria without proof; they are proved, like AA, by comparing the triangles with a smaller copy cut off by a parallel line. Their power is that you never need to measure an angle in SSS, and only one in SAS.
⚠️ Tests that do not work, and traps with the ones that do
1. Match sides by size in SSS. Put the sides of each triangle in order, shortest to longest, and divide the shortest by the shortest, the middle by the middle, the longest by the longest. Dividing in the order the sides happen to be named gives a wrong ratio.
2. The angle in SAS must be the included one. If the equal angle is not between the two proportional sides, the triangles need not be similar. "Two sides proportional and a non-included angle equal" (SSA) is not a criterion.
3. Two equal ratios of sides alone are not enough. Without a third ratio or an equal included angle, nothing is forced.
4. One equal angle alone is not enough. Two isosceles triangles with a $40^\circ$ angle may have apex $40^\circ$ in one and base angles $40^\circ$ in the other.
🧭 Problem-solving strategy — choosing the criterion
List what you are given: angles, side lengths, parallel lines, right angles, shared or vertically opposite angles.
Two angles? (Check for common angles, vertically opposite angles, angles from parallel lines, right angles.) Use AA.
Three sides? Sort them and compare the three ratios — SSS.
Two sides and the angle between them? Compare the two ratios and the included angle — SAS.
Write the similarity statement with matching vertices, list the three side ratios, and substitute the known lengths to find the unknown.
✏️ Example 3 — Using SSS
(a) $\triangle ABC$ has sides $4$, $6$, $8$ cm and $\triangle PQR$ has sides $6$, $9$, $12$ cm. Are they similar? (b) $\triangle ABC$ has sides $4$, $6$, $8$ cm and $\triangle XYZ$ has sides $5$, $7$, $9$ cm. Are they similar?
💡 Show solution▶
(a) Ratios Shortest : shortest $=\dfrac64=1.5$; middle : middle $=\dfrac96=1.5$; longest : longest $=\dfrac{12}{8}=1.5$. All three are equal.
(a) Conclude By SSS, the triangles are similar: with $AB=4$, $BC=6$, $CA=8$ and $PQ=6$, $QR=9$, $RP=12$ we get $\triangle ABC\sim\triangle PQR$, scale factor $1.5$.
(b) Ratios $\dfrac54=1.25$, $\dfrac76\approx1.167$, $\dfrac98=1.125$. They are not equal, so the triangles are not similar. (Adding $1$ to every side does not keep the shape.)
✏️ Example 4 — Using SAS
In $\triangle ABC$, $AB=6$ cm, $AC=9$ cm and $\angle A=48^\circ$. In $\triangle PQR$, $PQ=4$ cm, $PR=6$ cm and $\angle P=48^\circ$. Show that the triangles are similar, and find $BC$ if $QR=5$ cm.
💡 Show solution▶
Ratios $\dfrac{AB}{PQ}=\dfrac64=1.5$ and $\dfrac{AC}{PR}=\dfrac96=1.5$: equal.
Included angle $\angle A=\angle P=48^\circ$, and $\angle A$ lies between $AB$ and $AC$; $\angle P$ lies between $PQ$ and $PR$. By SAS, $\triangle ABC\sim\triangle PQR$.
Third side The scale factor is $1.5$, so $BC=1.5\times QR=1.5\times5=7.5$ cm.
Check $\dfrac{BC}{QR}=\dfrac{7.5}{5}=1.5$ ✓.
Interactive 1 — Build a Triangle from Only Some of the Parts
The blue triangle $ABC$ is fixed ($AB=8$, $BC=7$, $CA=5$, $\angle A=60^\circ$). Choose a criterion. The red triangle $DEF$ is then built using only the parts that criterion lists, with a size set by the slider $k$ — everything else is worked out from the picture. Watch the readouts: the remaining angles and all three side ratios come out equal automatically. The thick green arcs mark the angles you were given.
With the three criteria in hand, similar triangles appear everywhere: wherever there are parallel lines, a shared angle, a pair of right angles or a pair of vertically opposite angles, look for two triangles with a pair of equal angles. Then use the equal ratios to find lengths.
✏️ Example 5 — Overlapping triangles
In $\triangle ABC$, $D$ is on $AB$ and $E$ is on $AC$ with $\angle ADE=\angle ACB$. If $AB=15$ cm, $AC=10$ cm, $AD=6$ cm and $BC=10$ cm, find $AE$ and $DE$.
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Pick the triangles $\triangle ADE$ and $\triangle ACB$ share the angle $A$, and $\angle ADE=\angle ACB$ is given. By AA, $\triangle ADE\sim\triangle ACB$ with $A\leftrightarrow A$, $D\leftrightarrow C$, $E\leftrightarrow B$.
Ratios $\dfrac{AD}{AC}=\dfrac{AE}{AB}=\dfrac{DE}{CB}$, so $\dfrac{6}{10}=\dfrac{AE}{15}=\dfrac{DE}{10}$.
Solve $AE=\dfrac{6\times15}{10}=9$ cm and $DE=\dfrac{6\times10}{10}=6$ cm.
Check $E$ lies on $AC$ (as $9\lt10$ ✓). The ratio $0.6$ appears three times ✓. Note that $D$ matches $C$, not $B$: the order of the letters in the statement decides the ratios.
✏️ Example 6 — Two poles and crossing ropes
Two vertical poles $AB$ and $CD$ of heights $6$ m and $3$ m stand on level ground ($B$ and $D$ on the ground). A rope joins $A$ to $D$ and another joins $C$ to $B$; the ropes cross at $P$. Find the height of $P$ above the ground.
💡 Show solution▶
Set up Drop the perpendicular $PQ$ from $P$ to the ground $BD$ (so $Q$ is on $BD$). Then $PQ\parallel AB\parallel CD$ (all are vertical). Let $PQ=h$.
First pair In $\triangle DPQ$ and $\triangle DAB$: $\angle D$ is common and $\angle DQP=\angle DBA=90^\circ$, so $\triangle DPQ\sim\triangle DAB$ (AA). Hence $\dfrac{PQ}{AB}=\dfrac{DQ}{DB}$, i.e. $\dfrac{h}{6}=\dfrac{DQ}{DB}$.
Second pair In the same way $\triangle BPQ\sim\triangle BCD$, so $\dfrac{PQ}{CD}=\dfrac{BQ}{BD}$, i.e. $\dfrac{h}{3}=\dfrac{BQ}{BD}$.
Add $\dfrac h6+\dfrac h3=\dfrac{DQ+BQ}{DB}=\dfrac{DB}{DB}=1$. So $h\left(\dfrac16+\dfrac13\right)=\dfrac h2=1$ and $h=2$ m.
Check The answer does not depend on the distance $BD$ between the poles. In general the height is $\dfrac{ab}{a+b}$ for poles of heights $a$ and $b$: $\dfrac{6\times3}{9}=2$ ✓.
Interactive 2 — The SSS Tester: Similar or Not?
Pick a pair of triangles (or press Play to go through them all). The widget sorts the sides of each triangle from shortest to longest and divides: shortest ÷ shortest, middle ÷ middle, longest ÷ longest. If all three ratios agree, the triangles are similar. Use the slider to nudge the longest side of the red triangle and see how quickly one ratio breaks.
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blue sides (sorted)—
red sides (sorted)—
ratios—
SSS test—
Exercises
I
Are the triangles similar? Name the criterion and write the similarity statement when they are; otherwise say why the test fails.
6 Right triangles $ABC$ ($\angle A=90^\circ$, $\angle B=35^\circ$) and $PQR$ ($\angle P=90^\circ$, $\angle Q=35^\circ$).
7 Two isosceles triangles, each with apex angle $50^\circ$.
8 Sides $2$, $2$, $3$ cm and $4$, $4$, $7$ cm.
9 Right triangles $ABC$ and $PQR$ with $\angle A=\angle P=90^\circ$, $AB=3$, $AC=4$, $PQ=9$, $PR=12$ (cm).
10 Sides $7$, $8$, $9$ cm and $14$, $16$, $18$ cm.
💡 Show solutions▶
1 $\angle C=60^\circ$ and $\angle E=80^\circ$. Angles agree: similar by AA, $\triangle ABC\sim\triangle DEF$.
2 $\dfrac93=\dfrac{12}{4}=\dfrac{15}5=3$: similar by SSS.
3 $\dfrac{10}5=2$, $\dfrac{12}6=2$, $\dfrac{15}7\approx2.14$: not similar (the ratios differ).
4 $\dfrac64=\dfrac96=1.5$ with equal included angle: similar by SAS, $\triangle ABC\sim\triangle PQR$.
5 $\dfrac64=1.5$ but $\dfrac97\approx1.29$: ratios differ, so SAS fails; not similar.
6 $\angle A=\angle P$ and $\angle B=\angle Q$: similar by AA, $\triangle ABC\sim\triangle PQR$.
7 Base angles are $\dfrac{180^\circ-50^\circ}{2}=65^\circ$ in each, so all angles agree: similar by AA.
8 $\dfrac42=2$, $\dfrac42=2$, $\dfrac73\approx2.33$: not similar.
9 $\dfrac{PQ}{AB}=\dfrac93=3$, $\dfrac{PR}{AC}=\dfrac{12}{4}=3$, included angle $90^\circ$ each: similar by SAS.
10 $\dfrac{14}{7}=\dfrac{16}{8}=\dfrac{18}{9}=2$: similar by SSS.
II
Finding lengths.
1 $DE\parallel BC$ in $\triangle ABC$, $D$ on $AB$, $E$ on $AC$. $AD=4$ cm, $AB=10$ cm, $BC=15$ cm. Find $DE$.
2 $DE\parallel BC$, $AE=3$ cm, $AC=8$ cm and $DE=4.5$ cm. Find $BC$.
3 In $\triangle ABC$, $D$ is on $AB$, $E$ on $AC$, $\angle ADE=\angle ACB$. $AB=12$ cm, $AC=8$ cm, $AD=4$ cm, $BC=10$ cm. Find $AE$ and $DE$.
4 $AB\parallel CD$; $AC$ and $BD$ meet at $O$. Show $\triangle OAB\sim\triangle OCD$. If $OA=3$ cm, $OC=6$ cm and $AB=4$ cm, find $CD$.
5 $D$, $E$ are the midpoints of $AB$ and $AC$. Use SAS to show $\triangle ADE\sim\triangle ABC$ and deduce that $DE=\tfrac12BC$.
💡 Show solutions▶
1 $\triangle ADE\sim\triangle ABC$ (AA, Example 2): $\dfrac{DE}{BC}=\dfrac{AD}{AB}=\dfrac4{10}$, so $DE=0.4\times15=6$ cm.
2 $\dfrac{DE}{BC}=\dfrac{AE}{AC}=\dfrac38$, so $BC=\dfrac{4.5\times8}{3}=12$ cm.
3 $\triangle ADE\sim\triangle ACB$ (AA: common $\angle A$ and $\angle ADE=\angle ACB$), $D\leftrightarrow C$, $E\leftrightarrow B$. $\dfrac{AD}{AC}=\dfrac{AE}{AB}=\dfrac{DE}{CB}$: $\dfrac48=\dfrac{AE}{12}=\dfrac{DE}{10}=\dfrac12$. So $AE=6$ cm and $DE=5$ cm ($E$ is on $AC$ since $6\lt8$ ✓).
4 $\angle OAB=\angle OCD$ (alternate angles, $AB\parallel CD$), and $\angle AOB=\angle COD$ (vertically opposite): $\triangle OAB\sim\triangle OCD$ by AA. Then $\dfrac{AB}{CD}=\dfrac{OA}{OC}=\dfrac36=\dfrac12$, so $CD=8$ cm.
5 $\dfrac{AD}{AB}=\dfrac12=\dfrac{AE}{AC}$ and $\angle A$ is common (and included between those sides), so $\triangle ADE\sim\triangle ABC$ by SAS. Then $\dfrac{DE}{BC}=\dfrac12$, so $DE=\tfrac12BC$.
III
Applications. Justify each by naming the similar triangles and the criterion.
1 A girl $1.5$ m tall stands $2$ m behind a small mirror lying on the ground, $12$ m from a building. She sees the top of the building in the mirror. How tall is the building? (The ray makes equal angles with the ground on both sides of the mirror.)
2 To find the width $AB$ of a pond, a surveyor chooses a point $O$ with $OA=40$ m and $OB=60$ m, then marks $C$ on $AO$ extended with $OC=20$ m and $D$ on $BO$ extended with $OD=30$ m, and measures $CD=18$ m. Find $AB$.
3 A stick $1.2$ m tall casts a shadow $0.9$ m long. At the same time a tree's shadow is $7.5$ m long. How tall is the tree?
4 A triangular sail has sides $3$ m, $4$ m and $5$ m. A larger sail of the same shape has shortest side $4.5$ m. Find its other two sides.
5 A street lamp is $4.5$ m above the ground. A girl $1.5$ m tall stands $8$ m from the foot of the lamp. How long is her shadow?
6 Two poles $8$ m and $12$ m high stand on level ground and two ropes join the top of each pole to the foot of the other. Find the height of the point where the ropes cross.
7 A person's eye is $1.6$ m above the ground and $2.4$ m from a bowl of water on the ground. The bowl is $12$ m from the foot of a tree, and the person sees the top of the tree reflected in the water. Find the height of the tree.
💡 Show solutions▶
1 The two right triangles (girl–mirror–ray, building–mirror–ray) have a right angle and equal angles at the mirror: similar by AA. $\dfrac{h}{12}=\dfrac{1.5}{2}$, so $h=9$ m.
2 $\dfrac{OA}{OC}=\dfrac{40}{20}=2$ and $\dfrac{OB}{OD}=\dfrac{60}{30}=2$, and $\angle AOB=\angle COD$ (vertically opposite): $\triangle OAB\sim\triangle OCD$ by SAS. So $AB=2\times CD=36$ m.
3 Similar by AA (right angles, same Sun angle). $\dfrac{H}{7.5}=\dfrac{1.2}{0.9}=\dfrac43$, so $H=10$ m.
4 $k=\dfrac{4.5}{3}=1.5$ (SSS). The other sides are $4\times1.5=6$ m and $5\times1.5=7.5$ m.
5 Let the shadow be $x$ m. The triangles (lamp, ground to the shadow tip) and (girl, her shadow) are similar by AA with a shared angle at the shadow tip: $\dfrac{1.5}{4.5}=\dfrac{x}{x+8}$, so $x+8=3x$, $x=4$ m.
6 $h=\dfrac{ab}{a+b}=\dfrac{8\times12}{20}=4.8$ m (the result of Example 6, independent of the distance between the poles).
7 Right triangles with equal angles at the bowl (law of reflection): AA. $\dfrac{h}{12}=\dfrac{1.6}{2.4}=\dfrac23$, so $h=8$ m.
IV
Challenge.
1 $\triangle ABC\sim\triangle DEF$. $AM$ is the median to $BC$ (so $M$ is the midpoint of $BC$) and $DN$ is the median to $EF$. Prove that $\dfrac{AB}{DE}=\dfrac{AM}{DN}$.
2 $ABCD$ is a trapezium with $AB\parallel DC$ and $AB=2\,DC$. The diagonals meet at $O$. Show that $AO=2\,OC$.
3 $\triangle ABC$ and $\triangle PQR$ have $\angle A=\angle P$ and $\dfrac{AB}{PQ}=\dfrac{AC}{PR}=2$. If $AB=10$, $AC=12$ and $BC=14$ (cm), find the perimeter of $\triangle PQR$.
💡 Show solutions▶
1 From the similarity, $\dfrac{AB}{DE}=\dfrac{BC}{EF}$ and $\angle B=\angle E$. Since $BM=\tfrac12BC$ and $EN=\tfrac12EF$, $\dfrac{BM}{EN}=\dfrac{BC}{EF}=\dfrac{AB}{DE}$. So in $\triangle ABM$ and $\triangle DEN$: $\dfrac{AB}{DE}=\dfrac{BM}{EN}$ and $\angle B=\angle E$ (included), hence $\triangle ABM\sim\triangle DEN$ by SAS and $\dfrac{AM}{DN}=\dfrac{AB}{DE}$.
2 $\triangle OAB\sim\triangle OCD$ (AA: alternate angles and vertically opposite angles). So $\dfrac{AO}{OC}=\dfrac{AB}{DC}=2$, i.e. $AO=2\,OC$.
3 SAS gives $\triangle ABC\sim\triangle PQR$ with $k=2$ (from $PQR$ to $ABC$), so $QR=\dfrac{BC}{2}=7$ cm, $PQ=5$ cm, $PR=6$ cm. Perimeter $=5+6+7=18$ cm (half of $10+12+14=36$ cm).
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case study
How this is examined: 1-mark items ask which criterion applies or which similarity statement is correct; 2–3-mark problems ask you to prove two triangles similar by naming the criterion with its equal parts, then use the ratios to find a length; case studies use shadows, poles and lamp-posts. Always write the similarity statement in matching order and name the criterion — each earns a mark.
E1.Board · 1 mark In $\triangle ABC$ and $\triangle PQR$, $\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{CA}{RP}$. Then (A) $\triangle ABC\sim\triangle PQR$ (B) $\triangle ABC\sim\triangle QRP$ (C) $\triangle ABC\sim\triangle RPQ$ (D) $\triangle ABC\sim\triangle QPR$
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Criterion The three ratios are written with $AB\leftrightarrow PQ$, $BC\leftrightarrow QR$, $CA\leftrightarrow RP$, i.e. $A\leftrightarrow P$, $B\leftrightarrow Q$, $C\leftrightarrow R$. By SSS the triangles are similar in that order.
Trap The other options list the letters of $PQR$ in a different order, which would match $A$ with $Q$ or $R$ instead of $P$ — that pairs up different sides.
Answer (A).
E2.Board · 1 markAssertion (A): Any two right-angled triangles, each having an acute angle of $35^\circ$, are similar. Reason (R): If two angles of one triangle are equal to two angles of another triangle, the triangles are similar. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
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Check R This is the AA criterion (Theorem 6.3): true.
Check A Each triangle has a right angle ($90^\circ$) and an angle of $35^\circ$: two angles equal, so they are similar. The reason is exactly AA.
Answer (A).
E3.Board · 2 marks In $\triangle ABC$, $DE\parallel BC$ with $D$ on $AB$ and $E$ on $AC$. If $AD=6$ cm, $DB=9$ cm and $DE=4$ cm, find $BC$.
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Similar triangles $\angle A$ is common and $\angle ADE=\angle ABC$ (corresponding angles), so $\triangle ADE\sim\triangle ABC$ by AA. (1 mark)
Ratio $AB=6+9=15$ cm. $\dfrac{DE}{BC}=\dfrac{AD}{AB}=\dfrac{6}{15}=\dfrac25$, so $BC=\dfrac{4\times5}{2}=10$ cm. (1 mark)
E4.Board · 3 marks In $\triangle ABC$ and $\triangle PQR$, $\angle B=\angle Q=70^\circ$, $AB=6$ cm, $BC=9$ cm, $PQ=8$ cm and $QR=12$ cm. Show that $\triangle ABC\sim\triangle PQR$. If also $\angle A=45^\circ$, find $\angle R$.
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Ratios $\dfrac{AB}{PQ}=\dfrac68=\dfrac34$ and $\dfrac{BC}{QR}=\dfrac9{12}=\dfrac34$. (1 mark)
SAS The equal angles $\angle B=\angle Q$ are the included angles of these sides, so $\triangle ABC\sim\triangle PQR$ by SAS. (1 mark)
Angle $\angle C=180^\circ-45^\circ-70^\circ=65^\circ$, and $\angle R=\angle C=65^\circ$. (1 mark)
E5.Board · 3 marks In $\triangle ABC$, $D$ is a point on $BC$ such that $\angle ADC=\angle BAC$. Show that $CA^2=CB\times CD$. Hence find $CD$ if $CA=6$ cm and $CB=9$ cm.
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Similar triangles In $\triangle ADC$ and $\triangle BAC$: $\angle C$ is common and $\angle ADC=\angle BAC$ (given). By AA, $\triangle ADC\sim\triangle BAC$ with $A\leftrightarrow B$, $D\leftrightarrow A$, $C\leftrightarrow C$. (1 mark)
E6.Board · Case study A street lamp $AB$ stands on level ground ($B$ at its foot, $A$ at the bulb). A girl $CD$, $1.5$ m tall, stands $4$ m from the foot of the lamp ($BD=4$ m). Her shadow $DE$ (on the ground, away from the lamp) is $2$ m long. (i) Show that $\triangle EDC\sim\triangle EBA$, naming the criterion. (2 marks) (ii) Find the height of the lamp. (1 mark) (iii) The girl walks until she is $8$ m from the foot of the lamp. How long is her shadow now? (1 mark)
💡 Show solution▶
(i) Both $AB$ and $CD$ are vertical, so $\angle EDC=\angle EBA=90^\circ$. The angle at $E$ is common. By the AA criterion, $\triangle EDC\sim\triangle EBA$ ($E\leftrightarrow E$, $D\leftrightarrow B$, $C\leftrightarrow A$).
(ii) $BE=BD+DE=4+2=6$ m. $\dfrac{DC}{BA}=\dfrac{ED}{EB}$: $\dfrac{1.5}{AB}=\dfrac26$, so $AB=1.5\times3=4.5$ m.
(iii) Let the new shadow be $x$ m. $\dfrac{1.5}{4.5}=\dfrac{x}{x+8}$, so $x+8=3x$ and $x=4$ m. The shadow gets longer as she walks away from the lamp (from $2$ m to $4$ m).
🔬 Research–Teaching NexusMini project · optional, about 50 minutes
Research question. Can you measure the distance across an obstacle — a pond, a flower bed or a school field — without crossing it, using only a tape measure, string and similar triangles?
Why it matters. Surveyors measure the width of rivers, canyons and building plots without walking across them, and astronomers estimate the distance to the Moon and the planets from small triangles on Earth. The method is always the same: make a small triangle you can measure that is similar to a big triangle you cannot, and scale up. Errors in angles or lengths grow with the scale factor, so choosing a good scale factor is part of the skill.
What to do.
Choose two marks $A$ and $B$ on opposite sides of an obstacle (for example two trees on either side of the school field, or two chalk marks on opposite sides of a large puddle). You may walk around the obstacle but not across it.
Pick a point $O$ from which you can reach both $A$ and $B$ by tape. Measure $OA$ and $OB$.
On the lines $AO$ and $BO$ extended beyond $O$, mark points $C$ and $D$ with $OC=\tfrac12OA$ and $OD=\tfrac12OB$ (use string folded in half).
If you can, measure $AB$ directly on a different day to check. Repeat with a different point $O$ and with the ratio $\tfrac13$ instead of $\tfrac12$.
Tools: A long tape measure or measuring wheel, string, chalk or markers, a calculator
Hand in: A labelled sketch, a table with $OA$, $OB$, $OC$, $OD$, $CD$ and the calculated $AB$ for each attempt, the check measurement, your percentage error, and an explanation of why the method works.
Going further. If $O$ is so placed that $OA\ne OB$, does the method still work? What if $C$ and $D$ are marked on the same side of $O$ as $A$ and $B$ (between $O$ and $A$, $B$)? Draw the figure and decide which triangles are similar.
Teacher/parent note: this is the "crossed lines" configuration of Example 6 and Exercise III.2: $OC:OA=OD:OB$ with vertically opposite angles at $O$ gives similar triangles by SAS. A typical error is to measure $OC$ and $OD$ from the wrong point or to extend the lines not exactly in a straight line. A scale factor of $\tfrac12$ keeps the measuring errors small.
🧪 Self-check quiz
0/5 correct
Q1. Two triangles are similar if
Q2. Triangles with sides $3$, $5$, $7$ cm and $9$, $15$, $x$ cm are similar (shortest to shortest, etc.). Then $x=$
Q3. In $\triangle ABC$ and $\triangle DEF$, $\angle A=\angle D$ and $\dfrac{AB}{DE}=2$. Which extra fact proves $\triangle ABC\sim\triangle DEF$ by SAS?
Q4. $DE\parallel BC$ in $\triangle ABC$ ($D$ on $AB$, $E$ on $AC$), $AD=4$ cm, $AB=10$ cm, $BC=15$ cm. Then $DE=$
Q5. A vertical pole $2$ m high casts a shadow $3$ m long. At the same moment a tower casts a shadow $45$ m long. The height of the tower is