Basic Proportionality Theorem: Proof and Examples, Class 10

The Basic Pro­por­tion­al­ity The­o­rem#

An activ­ity that sug­gests the result#

Draw an angle at a point $A$. On one arm, mark off five equal steps and label the third and fifth marks $D$ and $B$. Then $AD$ is $3$ steps and $DB$ is $2$ steps, so $\displaystyle \frac{AD}{DB} = \frac{3}{2}$ just by count­ing. Through $B$ draw any line meet­ing the other arm at $C$. Through $D$ draw a line par­al­lel to $BC$, meet­ing $AC$ at $E$. Mea­sure $AE$ and $EC$: you will find $\displaystyle \frac{AE}{EC} = \frac{3}{2}$ as well, the same ratio as $\displaystyle \frac{AD}{DB}$.

That is no coin­ci­dence, and it gives us the the­o­rem.

State­ment#

The­o­rem 6.1 (Basic Pro­por­tion­al­ity The­o­rem). If a line is drawn par­al­lel to one side of a tri­an­gle to inter­sect the other two sides in dis­tinct points, the other two sides are divided in the same ratio.

In sym­bols: in $\triangle ABC$, if $D$ lies on $AB$, $E$ lies on $AC$ and $DE \parallel BC$, then

$$\displaystyle \frac{AD}{DB} = \frac{AE}{EC}.$$

Proof#

Given: $\triangle ABC$ with $D$ on $AB$, $E$ on $AC$ and $DE \parallel BC$.

To prove: $\displaystyle \frac{AD}{DB} = \frac{AE}{EC}$.

Con­struc­tion: Join $BE$ and $CD$. Draw $EN \perp AB$ (with $N$ on line $AB$) and $DM \perp AC$ (with $M$ on line $AC$).

Proof. The area of a tri­an­gle is $\displaystyle \frac{1}{2} \times \text{base} \times \text{height}$.

Step 1. Tri­an­gles $ADE$ and $BDE$ have bases $AD$ and $DB$ on the same line $AB$, and the same height $EN$. So

$$\displaystyle \frac{\text{ar}(ADE)}{\text{ar}(BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB}. \qquad (1)$$

Step 2. Tri­an­gles $ADE$ and $DEC$ have bases $AE$ and $EC$ on the same line $AC$, and the same height $DM$. So

$$\displaystyle \frac{\text{ar}(ADE)}{\text{ar}(DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}. \qquad (2)$$

Step 3. Tri­an­gles $BDE$ and $DEC$ stand on the same base $DE$ and lie between the same par­al­lels $DE$ and $BC$. Tri­an­gles on the same base and between the same par­al­lels have equal areas, so

$$\text{ar}(BDE) = \text{ar}(DEC). \qquad (3)$$

Step 4. By (3), the right-hand denom­i­na­tors in (1) and (2) are equal, and the numer­a­tors are the same area $\text{ar}(ADE)$. So the left-hand sides of (1) and (2) are equal:

$$\displaystyle \frac{AD}{DB} = \frac{AE}{EC}. \qquad \blacksquare$$

A use­ful rearrange­ment#

From $\displaystyle \frac{AD}{DB} = \frac{AE}{EC}$ we can also get the "whole side" forms

$$\displaystyle \frac{AD}{AB} = \frac{AE}{AC} \qquad \text{and} \qquad \frac{DB}{AB} = \frac{EC}{AC}.$$

Exam­ple 1 below shows how. These forms are handy when a ques­tion gives the whole side instead of the two pieces.

The con­verse of the the­o­rem#

State­ment#

If a line cuts two sides of a tri­an­gle in the same ratio, must it be par­al­lel to the third side? An activ­ity sug­gests it must. Mark off five equal steps on each arm of an angle at $A$ and join match­ing marks: the first mark on one arm to the first mark on the other, the sec­ond to the sec­ond, and so on. Every one of these join­ing lines divides the two arms in the same ratio, and every one comes out par­al­lel to the line join­ing the fifth marks.

The­o­rem 6.2 (Con­verse of the Basic Pro­por­tion­al­ity The­o­rem). If a line divides any two sides of a tri­an­gle in the same ratio, then the line is par­al­lel to the third side.

In sym­bols: in $\triangle ABC$, if $D$ is on $AB$, $E$ is on $AC$ and $\displaystyle \frac{AD}{DB} = \frac{AE}{EC}$, then $DE \parallel BC$.

Why the con­verse holds#

Sup­pose $\displaystyle \frac{AD}{DB} = \frac{AE}{EC}$ but $DE$ is not par­al­lel to $BC$. Draw $DE' \parallel BC$ with $E'$ on $AC$. By The­o­rem 6.1, $\displaystyle \frac{AD}{DB} = \frac{AE'}{E'C}$, so $\displaystyle \frac{AE}{EC} = \frac{AE'}{E'C}$. Adding $1$ to both sides gives $\displaystyle \frac{AC}{EC} = \frac{AC}{E'C}$, so $EC = E'C$. Then $E$ and $E'$ coin­cide, so $DE$ is the line $DE'$, which is par­al­lel to $BC$. This con­tra­dicts our assump­tion, so $DE \parallel BC$.

Together the two the­o­rems say: a line cut­ting two sides of a tri­an­gle is par­al­lel to the third side if and only if it divides those two sides in the same ratio.

Method: using the the­o­rem#

1. Find the tri­an­gle, and the line inside it that is par­al­lel to one side (or that you want to prove par­al­lel).
2. Name the com­mon ver­tex, the point where the two cut sides meet. Every ratio starts from that ver­tex: $\displaystyle \frac{\text{vertex to point}}{\text{point to far end}}$.
3. To find a length: write $\displaystyle \frac{AD}{DB} = \frac{AE}{EC}$, sub­sti­tute the three known val­ues, and solve.
4. To test for par­al­lel lines: com­pute both ratios (or cross-mul­ti­ply). Equal means par­al­lel; unequal means not par­al­lel.
5. For a proof involv­ing two tri­an­gles: find a ratio that both tri­an­gles share, and make it the link between them.

Worked exam­ples#

Exam­ple 1: the whole-side form#

A line inter­sects sides $AB$ and $AC$ of $\triangle ABC$ at $D$ and $E$ respec­tively and is par­al­lel to $BC$. Prove that $\displaystyle \frac{AD}{AB} = \frac{AE}{AC}$.

Solu­tion. Since $DE \parallel BC$, The­o­rem 6.1 gives $\displaystyle \frac{AD}{DB} = \frac{AE}{EC}$. Tak­ing rec­i­p­ro­cals, $\displaystyle \frac{DB}{AD} = \frac{EC}{AE}$. Add $1$ to both sides:

$$\displaystyle \begin{aligned} \frac{DB}{AD} + 1 &= \frac{EC}{AE} + 1 \\ \frac{DB + AD}{AD} &= \frac{EC + AE}{AE} \\ \frac{AB}{AD} &= \frac{AC}{AE} \end{aligned}$$

since $AD + DB = AB$ and $AE + EC = AC$. Tak­ing rec­i­p­ro­cals again, $\displaystyle \frac{AD}{AB} = \frac{AE}{AC}$. $\blacksquare$

Check with the fig­ure: $\displaystyle \frac{3}{5} = 0.6$ and $\displaystyle \frac{3.9}{6.5} = 0.6$.

Exam­ple 2: find­ing a length#

In $\triangle ABC$, $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$. If $AD = 2.4$ cm, $DB = 3.6$ cm and $AE = 2$ cm, find $EC$.

Solu­tion. By The­o­rem 6.1,

$$\displaystyle \frac{AD}{DB} = \frac{AE}{EC} \implies \frac{2.4}{3.6} = \frac{2}{EC} \implies EC = \frac{3.6 \times 2}{2.4} = 3.$$

Check: $\displaystyle \frac{2.4}{3.6} = \frac{2}{3}$ and $\displaystyle \frac{2}{3}$ match.

Answer: $EC = 3$ cm.

Exam­ple 3: an equa­tion from the the­o­rem#

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

Solu­tion. By The­o­rem 6.1,

$$\displaystyle \begin{aligned} \frac{x}{x - 2} &= \frac{x + 2}{x - 1} \\ x(x - 1) &= (x - 2)(x + 2) \\ x^2 - x &= x^2 - 4 \\ x &= 4 \end{aligned}$$

Check: $AD = 4$, $DB = 2$, $AE = 6$, $EC = 3$, and $\displaystyle \frac{4}{2} = \frac{6}{3} = 2$. All lengths are pos­i­tive, so the value is accept­able.

Answer: $x = 4$.

Exam­ple 4: is the line par­al­lel?#

$E$ and $F$ are points on sides $PQ$ and $PR$ of $\triangle PQR$. In each case, decide whether $EF \parallel QR$.

1. $PE = 3.9$ cm, $EQ = 3$ cm, $PF = 3.6$ cm, $FR = 2.4$ cm.
2. $PE = 4$ cm, $EQ = 4.5$ cm, $PF = 8$ cm, $FR = 9$ cm.
3. $PQ = 1.28$ cm, $PR = 2.56$ cm, $PE = 0.18$ cm, $PF = 0.36$ cm.

Solu­tion. We use The­o­rem 6.2.

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

(2) $\displaystyle \frac{PE}{EQ} = \frac{4}{4.5} = \frac{8}{9}$ and $\displaystyle \frac{PF}{FR} = \frac{8}{9}$. The ratios are equal, so $EF \parallel QR$.

(3) First find the pieces: $EQ = 1.28 - 0.18 = 1.10$ cm and $FR = 2.56 - 0.36 = 2.20$ cm. Then

$$\displaystyle \frac{PE}{EQ} = \frac{0.18}{1.10} = \frac{9}{55}, \qquad \frac{PF}{FR} = \frac{0.36}{2.20} = \frac{9}{55}.$$

The ratios are equal (both about $0.164$), so $EF \parallel QR$.

Answer: (1) not par­al­lel; (2) par­al­lel; (3) par­al­lel.

Exam­ple 5: a trapez­ium#

$ABCD$ is a trapez­ium with $AB \parallel DC$. $E$ and $F$ are points on the non-par­al­lel sides $AD$ and $BC$ respec­tively, such that $EF \parallel AB$. Show that $\displaystyle \frac{AE}{ED} = \frac{BF}{FC}$.

Solu­tion. Join $AC$ and let it meet $EF$ at $G$. Since $AB \parallel DC$ and $EF \parallel AB$, we also have $EF \parallel DC$ (lines par­al­lel to the same line are par­al­lel to each other).

In $\triangle ADC$: $E$ is on $AD$, $G$ is on $AC$ and $EG \parallel DC$. By The­o­rem 6.1,

$$\displaystyle \frac{AE}{ED} = \frac{AG}{GC}. \qquad (1)$$

In $\triangle CAB$: $G$ is on $CA$, $F$ is on $CB$ and $GF \parallel AB$. By The­o­rem 6.1 (ratios mea­sured from ver­tex $C$),

$$\displaystyle \frac{CG}{GA} = \frac{CF}{FB}, \quad \text{so} \quad \frac{AG}{GC} = \frac{BF}{FC}. \qquad (2)$$

From (1) and (2), $\displaystyle \frac{AE}{ED} = \frac{BF}{FC}$. $\blacksquare$

Exam­ple 6: prov­ing a tri­an­gle is isosce­les#

In $\triangle PQR$, $S$ and $T$ are points on $PQ$ and $PR$ such that $\displaystyle \frac{PS}{SQ} = \frac{PT}{TR}$ and $\angle PST = \angle PRQ$. Prove that $\triangle PQR$ is isosce­les.

Solu­tion. Since $\displaystyle \frac{PS}{SQ} = \frac{PT}{TR}$, The­o­rem 6.2 gives $ST \parallel QR$. With $PQ$ as the trans­ver­sal, $\angle PST$ and $\angle PQR$ are cor­re­spond­ing angles, so $\angle PST = \angle PQR$. We are also given $\angle PST = \angle PRQ$. There­fore $\angle PQR = \angle PRQ$. Sides oppo­site equal angles are equal, so $PR = PQ$, and $\triangle PQR$ is isosce­les. $\blacksquare$

Com­mon mis­takes#

• Mix­ing up the ratio: writ­ing $\displaystyle \frac{AD}{DB} = \frac{EC}{AE}$. Both ratios must run the same way from the com­mon ver­tex.
• Using $\displaystyle \frac{AD}{DB} = \frac{DE}{BC}$. The the­o­rem says noth­ing about $DE$ and $BC$ in this form; the cor­rect state­ment for them is $\displaystyle \frac{DE}{BC} = \frac{AD}{AB}$, which comes from sim­i­lar tri­an­gles.
• Apply­ing the the­o­rem when the line is not given (or proved) to be par­al­lel.
• For­get­ting to sub­tract when a whole side is given, as in Exam­ple 4 (3).
• Using The­o­rem 6.1 when a ques­tion asks you to prove lines par­al­lel. That needs the con­verse, The­o­rem 6.2.
• Accept­ing a value of $x$ that makes a length neg­a­tive.

Try these#

1. In $\triangle ABC$, $DE \parallel BC$, $AD = 3$ cm, $DB = 4.5$ cm and $AE = 2$ cm. Find $EC$. Answer: $3$ cm.
2. In $\triangle ABC$, $DE \parallel BC$, $AD = 6$ cm, $AB = 15$ cm and $AE = 4$ cm. Find $AC$. Answer: $10$ cm.
3. $PE = 5$ cm, $EQ = 4$ cm, $PF = 7.5$ cm, $FR = 6$ cm. Is $EF \parallel QR$? Answer: Yes.
4. $PE = 2$ cm, $EQ = 3$ cm, $PF = 3$ cm, $FR = 5$ cm. Is $EF \parallel QR$? Answer: No.
5. In $\triangle ABC$, $DE \parallel BC$ with $AD = 8x - 7$, $DB = 5x - 3$, $AE = 4x - 3$ and $EC = 3x - 1$. Find $x$. Answer: $x = 1$ (the other root, $\displaystyle x = -\tfrac{1}{2}$, gives neg­a­tive lengths).

Key terms#

Basic Pro­por­tion­al­ity The­o­rem

A line par­al­lel to one side of a tri­an­gle, cut­ting the other two sides in dis­tinct points, divides those sides in the same ratio.

Thales' The­o­rem

Another name for the Basic Pro­por­tion­al­ity The­o­rem, after Thales of Mile­tus.

Con­verse

The state­ment formed by swap­ping the hypoth­e­sis and the con­clu­sion of a the­o­rem.

Equian­gu­lar tri­an­gles

Tri­an­gles whose cor­re­spond­ing angles are all equal.

Divide in the same ratio

Two points cut two seg­ments so that the part-to-part ratios are equal.

Tri­an­gles between the same par­al­lels

Tri­an­gles whose bases lie on one line and whose oppo­site ver­tices lie on a par­al­lel line; on equal bases they have equal areas.

Trapez­ium

A quadri­lat­eral with at least one pair of par­al­lel sides.

Com­mon ques­tions#

Does the the­o­rem work if the line is par­al­lel to $AB$ instead of $BC$?#

Yes. It then cuts $CA$ and $CB$, and the ratios are mea­sured from ver­tex $C$. The com­mon ver­tex is always the one oppo­site the par­al­lel side.

Why does the proof use areas?#

Areas turn a state­ment about lengths into one about tri­an­gles with shared heights, and the par­al­lel lines give the key fact that $\text{ar}(BDE) = \text{ar}(DEC)$.

What is the dif­fer­ence between The­o­rem 6.1 and The­o­rem 6.2?#

The­o­rem 6.1 starts from par­al­lel lines and gives equal ratios. The­o­rem 6.2 starts from equal ratios and gives par­al­lel lines.

Is the mid-point the­o­rem a spe­cial case?#

Yes. If $D$ is the mid-point of $AB$ and $DE \parallel BC$, then $\displaystyle \frac{AE}{EC} = \frac{AD}{DB} = 1$, so $E$ is the mid-point of $AC$.

Can I cross-mul­ti­ply to test for par­al­lel lines?#

Yes. $\displaystyle \frac{PE}{EQ} = \frac{PF}{FR}$ exactly when $PE \times FR = PF \times EQ$, and this often avoids awk­ward dec­i­mals.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Heath, T. L. (trans.) The Thir­teen Books of Euclid's Ele­ments. Dover Pub­li­ca­tions.
3. Kise­lev, A. P. Kise­lev's Geom­e­try, Book I: Planime­try (adapted by A. Given­tal). Sum­iz­dat.
4. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.