Chapter 6 · Triangles

6.2 The Basic Proportionality Theorem

Draw a line parallel to one side of a triangle: it cuts the other two sides in the same ratio. This single fact — the Basic Proportionality Theorem — and its converse let you find lengths and prove lines parallel.

Look at the roof of a house from the side, or at a triangular set-square. A horizontal beam placed across the triangle, parallel to its base, seems to cut the two sloping sides "in proportion": if it is placed one-third of the way down one slope, it is one-third of the way down the other. This is not an accident — it is a theorem, one of the oldest in geometry, usually credited to the Greek mathematician Thales.

In this section we state the theorem carefully, prove it using areas (a Class 9 tool), and meet its converse, which lets us decide whether a line is parallel to a side just by measuring lengths along the other two sides. Together they give us a quick way to find unknown lengths, and the key to proving similarity of triangles in the next section.

Cutting a triangle with a parallel line

📐 Theorem 6.1 — Basic Proportionality Theorem (Thales)

Let $\triangle ABC$ be a triangle, and let $D$ be a point on side $AB$ and $E$ a point on side $AC$ (both different from the vertices). If $DE\parallel BC$, then $$\frac{AD}{DB}=\frac{AE}{EC}.$$ That is, a line drawn parallel to one side of a triangle, meeting the other two sides at two distinct points, divides those two sides in the same ratio.

Proof. Join $B$ to $E$ and $C$ to $D$. We compare areas, using $\text{area}=\tfrac12\times\text{base}\times\text{height}$.

Step 1. $\triangle ADE$ and $\triangle BDE$ have their bases $AD$ and $DB$ on the same line $AB$, and the same height — the perpendicular distance from $E$ to line $AB$. Triangles with equal heights have areas in the ratio of their bases: $$\frac{\text{area}(\triangle ADE)}{\text{area}(\triangle BDE)}=\frac{AD}{DB}.$$
Step 2. In the same way, $\triangle ADE$ and $\triangle CED$ have bases $AE$ and $EC$ on line $AC$ and the same height — the distance from $D$ to line $AC$. So $$\frac{\text{area}(\triangle ADE)}{\text{area}(\triangle CED)}=\frac{AE}{EC}.$$
Step 3. $\triangle BDE$ and $\triangle CED$ stand on the same base $DE$, and their other vertices $B$ and $C$ lie on the line $BC$, which is parallel to $DE$. So both have the same height (the distance between the parallel lines), and therefore equal areas: $\text{area}(\triangle BDE)=\text{area}(\triangle CED)$.
Step 4. The left-hand sides of Steps 1 and 2 have the same numerator, $\text{area}(\triangle ADE)$, and by Step 3 equal denominators. So the two left-hand sides are equal, and therefore so are the right-hand sides: $\dfrac{AD}{DB}=\dfrac{AE}{EC}$. $\blacksquare$

Other forms. From $\dfrac{AD}{DB}=\dfrac{AE}{EC}$, turn both fractions upside down: $\dfrac{DB}{AD}=\dfrac{EC}{AE}$, and add $1$ to each side: $\dfrac{DB+AD}{AD}=\dfrac{EC+AE}{AE}$. Since $AD+DB=AB$ and $AE+EC=AC$, this gives $$\frac{AB}{AD}=\frac{AC}{AE},\qquad\text{or equivalently}\qquad\frac{AD}{AB}=\frac{AE}{AC}\quad\text{and}\quad\frac{DB}{AB}=\frac{EC}{AC}.$$ Use whichever form has the numbers you know.

🧭 Problem-solving strategy — using the Basic Proportionality Theorem

Check the hypothesis: the line must be parallel to a side and its end points must lie on the other two sides. Mark the parallel sides on your sketch.
Identify the vertex $A$ where the two cut sides meet. The ratios are "part near $A$ : part far from $A$" on each side: $AD:DB$ and $AE:EC$.
Write one of the three forms of the proportion. Do not mix a "part : part" ratio on one side with a "part : whole" ratio on the other.
Solve for the unknown. Substitute back and check that both ratios agree; reject any value that makes a length negative.

✏️ Example 1 — Finding a missing length

In $\triangle ABC$, $D$ is on $AB$, $E$ is on $AC$ and $DE\parallel BC$. If $AD=3$ cm, $DB=5$ cm and $AE=4.5$ cm, find $EC$ and then the length $AC$.

💡 Show solution▶

Theorem Since $DE\parallel BC$, $\dfrac{AD}{DB}=\dfrac{AE}{EC}$ (Theorem 6.1).

Substitute $\dfrac35=\dfrac{4.5}{EC}$, so $EC=\dfrac{4.5\times5}{3}=7.5$ cm.

AC $AC=AE+EC=4.5+7.5=12$ cm.

Check $\dfrac{AE}{EC}=\dfrac{4.5}{7.5}=0.6=\dfrac35$ ✓. Also $\dfrac{AD}{AB}=\dfrac38$ and $\dfrac{AE}{AC}=\dfrac{4.5}{12}=\dfrac38$ ✓.

✏️ Example 2 — Lengths that contain an unknown

$DE\parallel BC$ in $\triangle ABC$ with $AD=x$ cm, $DB=(x-2)$ cm, $AE=(x+2)$ cm and $EC=(x-1)$ cm. Find $x$ and the four lengths.

💡 Show solution▶

Equation $\dfrac{AD}{DB}=\dfrac{AE}{EC}$ gives $\dfrac{x}{x-2}=\dfrac{x+2}{x-1}$.

Clear fractions $x(x-1)=(x-2)(x+2)$, so $x^2-x=x^2-4$.

Solve The $x^2$ terms cancel: $-x=-4$, so $x=4$.

Lengths $AD=4$, $DB=2$, $AE=6$, $EC=3$ (all in cm; all positive ✓).

Check $\dfrac42=2$ and $\dfrac63=2$ ✓.

The red segment $DE$ is always parallel to $BC$. Slide it down from $A$ with the slider (or press Play). Look at the ratios $\frac{AD}{DB}$ and $\frac{AE}{EC}$: although the lengths on the two sides are different, the two ratios are always equal. Tick Show the proof triangles to see why: $\triangle BDE$ (amber) and $\triangle CDE$ (green) always have the same area.

Position of $D$: $\frac{AD}{AB}=$0.40

$AD$, $DB$ (cm)—

$AE$, $EC$ (cm)—

$\tfrac{AD}{DB}$ and $\tfrac{AE}{EC}$—

area $\triangle BDE$ and $\triangle CDE$—

The converse: is the line parallel?

The theorem says: parallel $\Rightarrow$ equal ratios. Can we go the other way? If a line cuts two sides of a triangle in the same ratio, must it be parallel to the third side? Yes — and this is what lets us prove that lines are parallel by measuring lengths.

📐 Theorem 6.2 — Converse of the Basic Proportionality Theorem

Let $\triangle ABC$ be a triangle, with $D$ a point on side $AB$ and $E$ a point on side $AC$ (both different from the vertices). If $$\frac{AD}{DB}=\frac{AE}{EC},$$ then $DE\parallel BC$.

Why it is true. We give the idea, without the full formal write-up. Suppose $DE$ were not parallel to $BC$. Draw through $D$ the line parallel to $BC$, meeting $AC$ at a point $E'$ different from $E$. By Theorem 6.1, $\dfrac{AD}{DB}=\dfrac{AE'}{E'C}$. But we are told $\dfrac{AD}{DB}=\dfrac{AE}{EC}$, so $\dfrac{AE'}{E'C}=\dfrac{AE}{EC}$, which forces $E'$ and $E$ to be the same point (a point on a segment is fixed by the ratio in which it divides it). That contradicts $E\ne E'$. So $DE\parallel BC$.

⚠️ Three things to be careful about

1. Compare the right ratios. Test $\dfrac{AD}{DB}$ against $\dfrac{AE}{EC}$ (near part : far part on both sides), or $\dfrac{AD}{AB}$ against $\dfrac{AE}{AC}$ (part : whole on both). Never mix them.

2. Equal ratios on the wrong sides. If $AD:DB=AE:EC$ is false, the line is not parallel — it is enough to show the two ratios differ.

3. The theorem needs points on the sides. $D$ and $E$ must lie on the segments $AB$ and $AC$, with the lengths measured from the same vertex $A$.

✏️ Example 3 — Parallel or not?

In each case $E$ is on $PQ$ and $F$ is on $PR$ of $\triangle PQR$. Is $EF\parallel QR$?
(a) $PE=3.9$ cm, $EQ=3$ cm, $PF=3.6$ cm, $FR=2.4$ cm. (b) $PE=4$ cm, $EQ=4.5$ cm, $PF=8$ cm, $FR=9$ cm.

💡 Show solution▶

(a) $\dfrac{PE}{EQ}=\dfrac{3.9}{3}=1.3$ and $\dfrac{PF}{FR}=\dfrac{3.6}{2.4}=1.5$. The ratios differ, so $EF$ is not parallel to $QR$.

(b) $\dfrac{PE}{EQ}=\dfrac{4}{4.5}=\dfrac89$ and $\dfrac{PF}{FR}=\dfrac{8}{9}$. The ratios are equal, so by Theorem 6.2, $EF\parallel QR$.

Check (a) If $EF$ were parallel to $QR$, we would need $PF=1.3\times2.4=3.12$ cm, not $3.6$ cm.

✏️ Example 4 — Midpoints

(a) In $\triangle ABC$, $D$ is the midpoint of $AB$ and $DE\parallel BC$ meets $AC$ at $E$. Show that $E$ is the midpoint of $AC$.
(b) Conversely, if $D$ and $E$ are the midpoints of $AB$ and $AC$, show that $DE\parallel BC$.

💡 Show solution▶

(a) $D$ is the midpoint of $AB$, so $AD=DB$ and $\dfrac{AD}{DB}=1$. By Theorem 6.1, $\dfrac{AE}{EC}=1$, so $AE=EC$: $E$ is the midpoint of $AC$.

(b) $AD=DB$ and $AE=EC$, so $\dfrac{AD}{DB}=1=\dfrac{AE}{EC}$. By Theorem 6.2, $DE\parallel BC$.

Summary A line through the midpoint of one side, parallel to a second side, bisects the third side; and the segment joining the midpoints of two sides is parallel to the third side.

✏️ Example 5 — A fence in a triangular park

A triangular park $ABC$ has $AB=40$ m and $AC=56$ m. A straight fence $DE$ runs parallel to the road $BC$, with $D$ on $AB$ and $E$ on $AC$, and $AD=15$ m. Find $AE$ and $EC$.

💡 Show solution▶

Choose a form We know the whole sides, so use $\dfrac{AD}{AB}=\dfrac{AE}{AC}$: $\dfrac{15}{40}=\dfrac{AE}{56}$.

Solve $AE=\dfrac{15\times56}{40}=21$ m, so $EC=56-21=35$ m.

Check $\dfrac{AD}{DB}=\dfrac{15}{25}=\dfrac35$ and $\dfrac{AE}{EC}=\dfrac{21}{35}=\dfrac35$ ✓.

✏️ Example 6 — The diagonals of a trapezium

$ABCD$ is a trapezium with $AB\parallel DC$. Its diagonals $AC$ and $BD$ meet at $O$.
(a) Show that $\dfrac{AO}{OC}=\dfrac{BO}{OD}$.
(b) If $AO=(3x-1)$, $OC=(5x-3)$, $BO=(2x+1)$ and $OD=(6x-5)$ (all in cm), find $x$.

💡 Show solution▶

(a) Construction Through $O$ draw $EO\parallel AB$, meeting $AD$ at $E$. Since $AB\parallel DC$, also $EO\parallel DC$.

(a) Triangle ADC $EO\parallel DC$, with $E$ on $AD$ and $O$ on $AC$. By Theorem 6.1, $\dfrac{AE}{ED}=\dfrac{AO}{OC}$. (1)

(a) Triangle DAB $EO\parallel AB$, with $E$ on $DA$ and $O$ on $DB$. By Theorem 6.1, $\dfrac{DE}{EA}=\dfrac{DO}{OB}$, i.e. $\dfrac{AE}{ED}=\dfrac{OB}{OD}$. (2)

(a) Conclude From (1) and (2): $\dfrac{AO}{OC}=\dfrac{BO}{OD}$. $\blacksquare$

(b) Equation $\dfrac{3x-1}{5x-3}=\dfrac{2x+1}{6x-5}$, so $(3x-1)(6x-5)=(2x+1)(5x-3)$: $\;18x^2-21x+5=10x^2-x-3$, which gives $8x^2-20x+8=0$, i.e. $2x^2-5x+2=0$.

(b) Factorise $2x^2-5x+2=(2x-1)(x-2)=0$, so $x=\tfrac12$ or $x=2$.

(b) Reject For $x=\tfrac12$: $OC=5\left(\tfrac12\right)-3=-\tfrac12\lt0$, impossible for a length. So $x=2$, and $AO=5$, $OC=7$, $BO=5$, $OD=7$ cm.

Check $\dfrac57=\dfrac57$ ✓.

Now you choose both end points. Set $D$ on $AB$ and $E$ on $AC$ independently. The dashed green guide is the line through $D$ parallel to $BC$. Only when $\frac{AD}{DB}=\frac{AE}{EC}$ does $DE$ lie along the guide — and then the corresponding angles $\angle ADE$ and $\angle ABC$ are equal. Press Play to sweep $E$ and watch for the moment of parallelism.

$D$: $\frac{AD}{AB}=$0.40$E$: $\frac{AE}{AC}=$0.65

$\tfrac{AD}{DB}$—

$\tfrac{AE}{EC}$—

$\angle ADE$ and $\angle ABC$—

$DE\parallel BC$?—

Exercises

In each part $D$ is on $AB$, $E$ is on $AC$ of $\triangle ABC$ and $DE\parallel BC$.

1 $AD=2$ cm, $DB=3$ cm, $AE=4$ cm. Find $EC$.
2 $AD=4.5$ cm, $DB=3$ cm, $EC=5$ cm. Find $AE$.
3 $AB=12$ cm, $AD=8$ cm, $AC=15$ cm. Find $AE$ and $EC$.
4 $AD=(x+3)$ cm, $DB=x$ cm, $AE=(x+5)$ cm, $EC=(x+1)$ cm. Find $x$.
5 $\dfrac{AD}{DB}=\dfrac35$ and $AC=24$ cm. Find $AE$ and $EC$.
6 $AD=(4x-3)$, $DB=(3x-1)$, $AE=(8x-7)$, $EC=(5x-3)$ (in cm). Find $x$, rejecting any value that is impossible.
7 $AD=3$ cm, $AB=7.5$ cm, $AC=12$ cm. Find $AE$ and $EC$.
8 Starting from $\dfrac{AD}{DB}=\dfrac{AE}{EC}$, show that $\dfrac{AB}{AD}=\dfrac{AC}{AE}$.

💡 Show solutions▶

1 $\dfrac23=\dfrac4{EC}$, so $EC=\dfrac{4\times3}{2}=6$ cm.

2 $\dfrac{4.5}{3}=1.5=\dfrac{AE}{5}$, so $AE=7.5$ cm.

3 $\dfrac{AD}{AB}=\dfrac{8}{12}=\dfrac23=\dfrac{AE}{15}$, so $AE=10$ cm and $EC=15-10=5$ cm.

4 $(x+3)(x+1)=x(x+5)$: $x^2+4x+3=x^2+5x$, so $x=3$. Then $AD=6$, $DB=3$, $AE=8$, $EC=4$ (ratios $2$ and $2$ ✓).

5 $AE:EC=3:5$, so $AE=\dfrac38\times24=9$ cm and $EC=\dfrac58\times24=15$ cm.

6 $(4x-3)(5x-3)=(3x-1)(8x-7)$: $20x^2-27x+9=24x^2-29x+7$, so $4x^2-2x-2=0$, i.e. $2x^2-x-1=0=(2x+1)(x-1)$. So $x=1$ or $x=-\tfrac12$. For $x=-\tfrac12$ the length $AD=4\left(-\tfrac12\right)-3=-5$ is negative: reject. $x=1$: $AD=1$, $DB=2$, $AE=1$, $EC=2$ ✓.

7 $\dfrac{AD}{AB}=\dfrac{3}{7.5}=0.4=\dfrac{AE}{12}$, so $AE=4.8$ cm and $EC=12-4.8=7.2$ cm.

8 Turn each fraction upside down: $\dfrac{DB}{AD}=\dfrac{EC}{AE}$. Add $1$ to both sides: $\dfrac{DB+AD}{AD}=\dfrac{EC+AE}{AE}$. Since $DB+AD=AB$ and $EC+AE=AC$, we get $\dfrac{AB}{AD}=\dfrac{AC}{AE}$.

The converse. $D$ lies on $AB$ and $E$ on $AC$ of $\triangle ABC$ (or the corresponding sides in the named triangle). In which cases is the line parallel to the third side?

1 In $\triangle PQR$: $PE=3.9$, $EQ=3$, $PF=3.6$, $FR=2.4$ (cm), $E$ on $PQ$, $F$ on $PR$. Is $EF\parallel QR$?
2 In $\triangle PQR$: $PE=4$, $EQ=4.5$, $PF=8$, $FR=9$ (cm). Is $EF\parallel QR$?
3 $AD=8$, $AB=12$, $AE=12$, $AC=18$ (cm). Is $DE\parallel BC$?
4 $AD=3$, $AB=9$, $AE=5$, $AC=12$ (cm). Is $DE\parallel BC$?
5 Show that the line segment joining the midpoints of two sides of a triangle is parallel to the third side.
6 Find $x$ so that $DE\parallel BC$, if $AD=x$, $DB=(x+2)$, $AE=3$, $EC=5$ (all in cm).
7 In $\triangle ABC$, $\dfrac{AD}{DB}=\dfrac{AE}{EC}=\dfrac35$ ($D$ on $AB$, $E$ on $AC$). If $\angle ADE=52^\circ$, find $\angle ABC$. Give the reason.

💡 Show solutions▶

1 $\dfrac{3.9}{3}=1.3$ and $\dfrac{3.6}{2.4}=1.5$: different, so not parallel.

2 $\dfrac{4}{4.5}=\dfrac89$ and $\dfrac{8}{9}$: equal, so parallel (Theorem 6.2).

3 $DB=12-8=4$ and $EC=18-12=6$. $\dfrac{AD}{DB}=\dfrac84=2$, $\dfrac{AE}{EC}=\dfrac{12}{6}=2$: parallel.

4 $DB=6$, $EC=7$. $\dfrac{AD}{DB}=\dfrac36=0.5$, $\dfrac{AE}{EC}=\dfrac57\approx0.714$: different, not parallel.

5 Let $D$, $E$ be the midpoints of $AB$, $AC$. Then $AD=DB$ and $AE=EC$, so $\dfrac{AD}{DB}=1=\dfrac{AE}{EC}$. By Theorem 6.2, $DE\parallel BC$.

6 $\dfrac{x}{x+2}=\dfrac35$, so $5x=3x+6$, $x=3$. Then $AD=3$, $DB=5$ and $\dfrac35=\dfrac35$ ✓.

7 The ratios are equal, so $DE\parallel BC$ (Theorem 6.2). $\angle ADE$ and $\angle ABC$ are corresponding angles on the parallel lines $DE$ and $BC$ with transversal $AB$, so $\angle ABC=52^\circ$.

III

Applications and proofs.

1 A triangular plot $ABC$ has $AB=40$ m and $AC=56$ m. A path $DE\parallel BC$ is laid with $D$ on $AB$, $E$ on $AC$ and $AD=15$ m. Find $AE$ and $EC$ — then the plot is re-planned with $AD=24$ m: find $AE$.
2 In a trapezium $ABCD$ with $AB\parallel DC$ the diagonals meet at $O$. If $AO=6$ cm, $OC=4$ cm and $BO=9$ cm, find $OD$.
3 $D$ is the midpoint of $AB$, $DE\parallel BC$ and $AC=11.6$ cm. Find $AE$.
4 A roof frame $ABC$ has equal sloping sides $AB=AC=6.5$ m. A horizontal beam $DE\parallel BC$ is fixed with $AD=AE=2.6$ m. Find $DB$ and $EC$. Is it true that the beam divides both slopes in the same ratio?
5 $DE\parallel BC$, $\dfrac{AD}{DB}=\dfrac37$ and $AC=14$ cm. Find $AE$ and $EC$.
6 Prove: in a trapezium $ABCD$ with $AB\parallel DC$, the line through the midpoint $E$ of $AD$ parallel to $AB$ passes through the midpoint of $BC$.
7 The diagonals of a quadrilateral $ABCD$ meet at $O$ and $\dfrac{AO}{OC}=\dfrac{BO}{OD}$. Show that $ABCD$ is a trapezium ($AB\parallel DC$).
8 In the trapezium $ABCD$ ($AB\parallel DC$), $AO=(x+1)$, $OC=2x$, $BO=(3x-3)$, $OD=(4x-3)$ (cm). Find $x$, rejecting impossible values, and the four lengths.

💡 Show solutions▶

1 $\dfrac{AD}{AB}=\dfrac{AE}{AC}$: $AE=\dfrac{15\times56}{40}=21$ m, $EC=35$ m. With $AD=24$: $AE=\dfrac{24\times56}{40}=33.6$ m (and $EC=22.4$ m).

2 $\dfrac{AO}{OC}=\dfrac{BO}{OD}$ (Example 6): $\dfrac64=\dfrac9{OD}$, so $OD=\dfrac{9\times4}{6}=6$ cm.

3 $\dfrac{AD}{DB}=1$ so $E$ is the midpoint of $AC$ (Example 4): $AE=11.6\div2=5.8$ cm.

4 $DB=6.5-2.6=3.9$ m and $EC=3.9$ m. $\dfrac{AD}{DB}=\dfrac{2.6}{3.9}=\dfrac23$ and $\dfrac{AE}{EC}=\dfrac23$: yes, equal ratios, so by Theorem 6.2 the beam is indeed parallel to the base.

5 $AE:EC=3:7$, so $AE=\dfrac{3}{10}\times14=4.2$ cm and $EC=9.8$ cm.

6 Let $EF\parallel AB$ meet $BC$ at $F$, and join $AC$, meeting $EF$ at $G$. In $\triangle ADC$: $EG\parallel DC$ (since $EF\parallel AB\parallel DC$) and $\dfrac{AE}{ED}=1$, so $\dfrac{AG}{GC}=1$: $G$ is the midpoint of $AC$. In $\triangle CAB$: $GF\parallel AB$, and $\dfrac{CG}{GA}=1$, so by Theorem 6.1 $\dfrac{CF}{FB}=1$. Hence $F$ is the midpoint of $BC$.

7 Through $O$ draw $OE\parallel AB$ meeting $AD$ at $E$. In $\triangle DAB$: $\dfrac{DE}{EA}=\dfrac{DO}{OB}$. From the given ratio, $\dfrac{DO}{OB}=\dfrac{OC}{AO}$ (turn the equation $\frac{AO}{OC}=\frac{BO}{OD}$ upside down and rearrange: $\frac{OD}{BO}=\frac{OC}{AO}$). So $\dfrac{DE}{EA}=\dfrac{OC}{AO}$, i.e. $\dfrac{AE}{ED}=\dfrac{AO}{OC}$. By Theorem 6.2 in $\triangle ADC$, $EO\parallel DC$. Since also $EO\parallel AB$, we get $AB\parallel DC$, and $ABCD$ is a trapezium.

8 $\dfrac{x+1}{2x}=\dfrac{3x-3}{4x-3}$, so $(x+1)(4x-3)=2x(3x-3)$: $4x^2+x-3=6x^2-6x$, giving $2x^2-7x+3=0=(2x-1)(x-3)$. So $x=\tfrac12$ or $x=3$. For $x=\tfrac12$, $BO=3(\tfrac12)-3=-1.5\lt0$: reject. $x=3$: $AO=4$, $OC=6$, $BO=6$, $OD=9$ cm, and $\tfrac46=\tfrac69$ ✓.

Challenge — chains of parallel lines.

1 In $\triangle ABC$, $DE\parallel BC$ ($D$ on $AB$, $E$ on $AC$) and $\dfrac{AD}{DB}=\dfrac23$. Find $\dfrac{AE}{AC}$.
2 In $\triangle ABC$, $DE\parallel BC$ with $D$ on $AB$ and $E$ on $AC$. A point $F$ on $AD$ is such that $EF\parallel CD$. Prove that $AD^2=AF\times AB$.
3 In the setting of Part 2, $AB=16$ cm and $AD=8$ cm. Find $AF$ and check Part 2.

💡 Show solutions▶

1 $\dfrac{AE}{EC}=\dfrac23$, so $AE:AC=2:5$ and $\dfrac{AE}{AC}=\dfrac25$.

2 $DE\parallel BC$ in $\triangle ABC$ gives $\dfrac{AD}{AB}=\dfrac{AE}{AC}$. $EF\parallel CD$ in $\triangle ACD$ (with $F$ on $AD$, $E$ on $AC$) gives $\dfrac{AF}{AD}=\dfrac{AE}{AC}$. So $\dfrac{AD}{AB}=\dfrac{AF}{AD}$, hence $AD^2=AF\times AB$.

3 $AD^2=64=AF\times16$, so $AF=4$ cm. Check: $\dfrac{AF}{AD}=\dfrac48=\dfrac12=\dfrac{AD}{AB}$ ✓.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · 5 marks

How this is examined: the Basic Proportionality Theorem is a standard 5-mark "state and prove" question (statement, figure with construction, and three or four reasoned steps). Short questions use it to find a missing length or an unknown $x$ (2–3 marks) and the converse to decide parallelism (1–3 marks). Always name the theorem in each line you use it.

E1. Board · 1 mark In $\triangle ABC$, $DE\parallel BC$ with $D$ on $AB$ and $E$ on $AC$. If $AD=2$ cm, $DB=3$ cm and $AE=3$ cm, then $EC=$ (A) $2$ cm (B) $3$ cm (C) $4.5$ cm (D) $6$ cm

💡 Show solution▶

Theorem $\dfrac{AD}{DB}=\dfrac{AE}{EC}$, so $\dfrac23=\dfrac{3}{EC}$ and $EC=\dfrac{3\times3}{2}=4.5$ cm.

Trap (B) $3$ cm just copies $AE$; (D) $6$ cm comes from multiplying $3\times2$ instead of working out $\dfrac{3\times3}{2}$ — it inverts the ratio. The far part $EC$ must be bigger than $AE$ because $DB\gt AD$.

Answer (C).

E2. Board · 1 mark Assertion (A): In $\triangle ABC$, $D$ and $E$ are points on $AB$ and $AC$ with $AD=3$ cm, $DB=4$ cm, $AE=6$ cm, $EC=8$ cm. Then $DE\parallel BC$. Reason (R): If a line divides two sides of a triangle in the same ratio, then it is parallel to the third side.
(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.

💡 Show solution▶

Check R This is Theorem 6.2, the converse of the Basic Proportionality Theorem. True.

Check A $\dfrac{AD}{DB}=\dfrac34$ and $\dfrac{AE}{EC}=\dfrac68=\dfrac34$: equal, so $DE\parallel BC$. True, and the reason is exactly R.

Answer (A).

E3. Board · 2 marks In $\triangle ABC$, $DE\parallel BC$ with $D$ on $AB$ and $E$ on $AC$. If $AD=x$ cm, $DB=(3x+4)$ cm, $AE=(x+3)$ cm and $EC=(3x+19)$ cm, find $x$.

💡 Show solution▶

Theorem $\dfrac{AD}{DB}=\dfrac{AE}{EC}$: $\dfrac{x}{3x+4}=\dfrac{x+3}{3x+19}$. (1 mark)

Solve $x(3x+19)=(3x+4)(x+3)$: $3x^2+19x=3x^2+13x+12$, so $6x=12$ and $x=2$. (1 mark) Check: $AD=2$, $DB=10$, $AE=5$, $EC=25$; $\tfrac{2}{10}=\tfrac{5}{25}$ ✓.

E4. Board · 3 marks In $\triangle ABC$, $D$ is on $AB$ and $E$ on $AC$ with $AB=12$ cm, $AD=8$ cm, $AC=21$ cm and $AE=14$ cm. Show that $DE\parallel BC$, and find $\angle ADE$ if $\angle ABC=65^\circ$.

💡 Show solution▶

Lengths $DB=12-8=4$ cm and $EC=21-14=7$ cm. (1 mark)

Converse $\dfrac{AD}{DB}=\dfrac84=2$ and $\dfrac{AE}{EC}=\dfrac{14}{7}=2$. The ratios are equal, so by the converse of the Basic Proportionality Theorem, $DE\parallel BC$. (1 mark)

Angle $\angle ADE$ and $\angle ABC$ are corresponding angles for parallel lines $DE$, $BC$ and transversal $AB$, so $\angle ADE=\angle ABC=65^\circ$. (1 mark)

E5. Board · 5 marks State and prove the Basic Proportionality Theorem.

💡 Show solution▶

Statement (1 mark) If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, it divides those two sides in the same ratio.

Given, to prove (1 mark) In $\triangle ABC$, $D$ on $AB$, $E$ on $AC$, $DE\parallel BC$. To prove: $\dfrac{AD}{DB}=\dfrac{AE}{EC}$. Construction: join $BE$ and $CD$, and draw $EF\perp AB$, $DG\perp AC$.

Proof (3 marks) $\dfrac{\text{ar}(ADE)}{\text{ar}(BDE)}=\dfrac{\frac12\times AD\times EF}{\frac12\times DB\times EF}=\dfrac{AD}{DB}$ and $\dfrac{\text{ar}(ADE)}{\text{ar}(CED)}=\dfrac{\frac12\times AE\times DG}{\frac12\times EC\times DG}=\dfrac{AE}{EC}$. Triangles $BDE$ and $CED$ are on the same base $DE$ and between the same parallels $DE$ and $BC$, so $\text{ar}(BDE)=\text{ar}(CED)$. Hence $\dfrac{AD}{DB}=\dfrac{AE}{EC}$. $\blacksquare$

🔬 Research–Teaching NexusMini project · optional, about 40 minutes

Research question. Can you divide a strip of paper or a ribbon into three equal parts, or into two parts in the ratio $2:3$, using only the ruled lines of a notebook page and no measuring at all?

Why it matters. Tailors, carpenters and architects often need to divide a length into equal or proportional parts when the numbers do not divide neatly — for example, spacing five equal stairs along a slope, or placing shelves evenly on a wall. The trick is to use parallel lines, and the reason it works is the Basic Proportionality Theorem: parallel lines cut transversals in proportion. The same idea is built into the "ruled-paper" and "proportional divider" tools used by draughtsmen.

What to do.

Take a ruled sheet (the lines are parallel and equally spaced) and a strip of paper with its two ends marked $A$ and $B$. Choose any length, say $10$ cm.
Lay the strip across the page so that end $A$ sits exactly on the first ruled line and end $B$ sits exactly on the fourth ruled line. Mark with a pencil where the strip crosses the 2nd and 3rd lines.
Measure the three parts with a ruler. How close to $\tfrac{10}{3}\approx3.33$ cm are they?
Now divide the strip in the ratio $2:3$: place $A$ on line 1 and $B$ on line 6 (five gaps) and mark where it crosses line 3.
Draw a neat sketch: the strip, the ruled lines, and the marked points. Explain in a few sentences why the points are equally spaced (hint: through $A$ draw a line parallel to the transversal and use Example 6 or the midpoint idea).

Tools: A ruled notebook page, a strip of paper or ribbon, a pencil, a ruler

Hand in: Your marked strips, a table of the measured lengths against the expected lengths with the percentage error, and your explanation.

Going further. Divide a $14$ cm strip into $7$ equal parts this way. What goes wrong if the lines on the page are not exactly parallel — try a page where the lines are drawn freehand and compare.

Teacher/parent note: the property used is that parallel lines cutting off equal intercepts on one transversal cut off equal intercepts on every transversal — a consequence of the Basic Proportionality Theorem (draw through each division point a line parallel to the other transversal and use congruent triangles, or the midpoint result of Example 4). Student errors usually come from the strip not touching the lines exactly at its end points.

🧪 Self-check quiz

0/5 correct

Q1. In $\triangle ABC$, $DE\parallel BC$ ($D$ on $AB$, $E$ on $AC$). If $AD=3$ cm, $DB=4$ cm and $AE=4.5$ cm, then $EC=$

Q2. If $DE\parallel BC$ in $\triangle ABC$ ($D$ on $AB$, $E$ on $AC$), which equation is correct?

Q3. $D$ is on $AB$ and $E$ on $AC$ of $\triangle ABC$. In which case is $DE\parallel BC$?

Q4. $D$ is the midpoint of $AB$, $DE\parallel BC$ with $E$ on $AC$, and $AC=9.2$ cm. Then $AE=$

Q5. $DE\parallel BC$, $AB=10$ cm, $AD=4$ cm and $AC=15$ cm. Then $EC=$