NCERT Class 10 Mathematics — Triangles, Exercise 6.2. All 10 questions solved.

Exercise 6.2 runs on one theorem and its converse.

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

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

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

Inverting the first ratio and adding $1$ to both sides gives $\dfrac{AB}{AD} = \dfrac{AC}{AE}$, so the theorem can equally be used in the part-to-whole form

$$\frac{AD}{AB} = \frac{AE}{AC}.$$

What must never happen is a part-to-part ratio on one side of the equation and a part-to-whole ratio on the other.

Key insight. The two theorems run in opposite directions. Theorem 6.1 turns parallel lines into equal ratios; Theorem 6.2 turns equal ratios back into parallel lines. Every proof from question 3 onwards is the same chain: a parallel line in one triangle gives a ratio, a parallel line in a second triangle sharing a side gives a ratio with the same term on that shared side, and the common term ties the two together. When the thing to be proved is itself a parallel line (questions 5, 6, 8 and 10), the chain ends with the converse.

Question 1

In Fig. 6.17, (i) and (ii), $DE \parallel BC$. Find $EC$ in (i) and $AD$ in (ii).

In both parts $D$ lies on $AB$ and $E$ on $AC$. In (i), $AD = 1.5$ cm, $DB = 3$ cm and $AE = 1$ cm. In (ii), $DB = 7.2$ cm, $AE = 1.8$ cm and $EC = 5.4$ cm.

Solution. In both parts $DE \parallel BC$, so the Basic Proportionality Theorem applies to $\triangle ABC$:

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

Three of the four lengths are given each time, so the fourth follows.

(i)

$$\frac{1.5}{3} = \frac{1}{EC} \quad\Longrightarrow\quad EC = \frac{3 \times 1}{1.5} = 2 \text{ cm}.$$

Both ratios are $1 : 2$, so $E$ divides $AC$ exactly as $D$ divides $AB$.

(ii)

$$\frac{AD}{7.2} = \frac{1.8}{5.4} = \frac{1}{3} \quad\Longrightarrow\quad AD = \frac{7.2}{3} = 2.4 \text{ cm}.$$

Question 2

$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle PQR$. For each of the following cases, state whether $EF \parallel QR$:

(i) $PE = 3.9$ cm, $EQ = 3$ cm, $PF = 3.6$ cm and $FR = 2.4$ cm

(ii) $PE = 4$ cm, $QE = 4.5$ cm, $PF = 8$ cm and $RF = 9$ cm

(iii) $PQ = 1.28$ cm, $PR = 2.56$ cm, $PE = 0.18$ cm and $PF = 0.36$ cm

Solution. Together the two theorems say that $EF \parallel QR$ exactly when $E$ and $F$ divide $PQ$ and $PR$ in the same ratio. So compute $\dfrac{PE}{EQ}$ and $\dfrac{PF}{FR}$ and compare them.

(i) $\dfrac{PE}{EQ} = \dfrac{3.9}{3} = 1.3$ and $\dfrac{PF}{FR} = \dfrac{3.6}{2.4} = 1.5$. The ratios differ. If $EF$ were parallel to $QR$, the Basic Proportionality Theorem would force them to be equal, so $EF$ is not parallel to $QR$.

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

(iii) Here whole sides are given, so find the remaining pieces first: $EQ = PQ – PE = 1.28 – 0.18 = 1.10$ cm and $FR = PR – PF = 2.56 – 0.36 = 2.20$ cm. Then

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

The ratios are equal, so $EF \parallel QR$. Comparing the part-to-whole ratios $\dfrac{PE}{PQ} = \dfrac{0.18}{1.28} = \dfrac{9}{64}$ and $\dfrac{PF}{PR} = \dfrac{0.36}{2.56} = \dfrac{9}{64}$ gives the same conclusion without the subtraction.

Question 3

In Fig. 6.18, if $LM \parallel CB$ and $LN \parallel CD$, prove that $\dfrac{AM}{AB} = \dfrac{AN}{AD}$.

The figure is a quadrilateral $ABCD$ with its diagonal $AC$ drawn. $M$ lies on $AB$, $L$ on $AC$ and $N$ on $AD$.

Solution. The diagonal $AC$ splits the figure into $\triangle ABC$ and $\triangle ADC$, and each triangle contains one of the parallel lines. Both ratios we want can then be tied to the same ratio on the shared side $AC$.

In $\triangle ABC$, $LM \parallel CB$, so by the Basic Proportionality Theorem

$$\frac{AM}{MB} = \frac{AL}{LC}. \qquad (1)$$

In $\triangle ADC$, $LN \parallel CD$, so

$$\frac{AN}{ND} = \frac{AL}{LC}. \qquad (2)$$

From (1) and (2), $\dfrac{AM}{MB} = \dfrac{AN}{ND}$. The question wants the whole sides $AB$ and $AD$ in the denominators, so invert both sides and add $1$:

$$\frac{MB}{AM} + 1 = \frac{ND}{AN} + 1 \quad\Longrightarrow\quad \frac{MB + AM}{AM} = \frac{ND + AN}{AN} \quad\Longrightarrow\quad \frac{AB}{AM} = \frac{AD}{AN}.$$

Inverting once more gives the result.

Question 4

In Fig. 6.19, $DE \parallel AC$ and $DF \parallel AE$. Prove that $\dfrac{BF}{FE} = \dfrac{BE}{EC}$.

In the figure, $D$ is on $AB$, and $F$ and $E$ are on $BC$ with $F$ between $B$ and $E$. The segments $AE$, $DE$ and $DF$ are drawn.

Solution. Look for two triangles with vertex $B$ that each contain one of the parallel lines. They share the side $BA$, so the ratio $\dfrac{BD}{DA}$ will link them.

In $\triangle ABE$, $DF \parallel AE$, with $D$ on $BA$ and $F$ on $BE$. By the Basic Proportionality Theorem,

$$\frac{BD}{DA} = \frac{BF}{FE}. \qquad (1)$$

In $\triangle ABC$, $DE \parallel AC$, with $D$ on $BA$ and $E$ on $BC$, so

$$\frac{BD}{DA} = \frac{BE}{EC}. \qquad (2)$$

The left-hand sides of (1) and (2) are the same, so the right-hand sides are equal.

Question 5

In Fig. 6.20, $DE \parallel OQ$ and $DF \parallel OR$. Show that $EF \parallel QR$.

$O$ is a point inside $\triangle PQR$, joined to $P$, $Q$ and $R$. $D$ lies on $PO$, $E$ on $PQ$ and $F$ on $PR$.

Solution. To prove that two lines are parallel we need the converse, so the aim is to show that $E$ and $F$ divide $PQ$ and $PR$ in the same ratio. The ratio $\dfrac{PD}{DO}$ on the shared side $PO$ will link them.

In $\triangle POQ$, $DE \parallel OQ$, so

$$\frac{PE}{EQ} = \frac{PD}{DO}. \qquad (1)$$

In $\triangle POR$, $DF \parallel OR$, so

$$\frac{PF}{FR} = \frac{PD}{DO}. \qquad (2)$$

From (1) and (2), $\dfrac{PE}{EQ} = \dfrac{PF}{FR}$. So in $\triangle PQR$ the line $EF$ divides the sides $PQ$ and $PR$ in the same ratio, and by the converse of the Basic Proportionality Theorem (Theorem 6.2), $EF \parallel QR$.

Question 6

In Fig. 6.21, $A$, $B$ and $C$ are points on $OP$, $OQ$ and $OR$ respectively such that $AB \parallel PQ$ and $AC \parallel PR$. Show that $BC \parallel QR$.

$O$ is a point inside $\triangle PQR$, joined to $P$, $Q$ and $R$, and $\triangle ABC$ sits inside it with its vertices on $OP$, $OQ$ and $OR$.

Solution. This follows the pattern of question 5, with the shared ratio now on $OP$.

In $\triangle OPQ$, $AB \parallel PQ$, so

$$\frac{OA}{AP} = \frac{OB}{BQ}. \qquad (1)$$

In $\triangle OPR$, $AC \parallel PR$, so

$$\frac{OA}{AP} = \frac{OC}{CR}. \qquad (2)$$

From (1) and (2), $\dfrac{OB}{BQ} = \dfrac{OC}{CR}$. In $\triangle OQR$, then, $B$ and $C$ divide $OQ$ and $OR$ in the same ratio, and by the converse of the Basic Proportionality Theorem, $BC \parallel QR$.

Question 7

Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Solution. Let $D$ be the mid-point of side $AB$ of $\triangle ABC$, and let the line through $D$ parallel to $BC$ meet $AC$ at $E$. We must show that $AE = EC$.

The starting point is a parallel line, so Theorem 6.1 is the tool. Since $DE \parallel BC$,

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

$D$ is the mid-point of $AB$, so $AD = DB$ and the left-hand side is $1$. Hence $\dfrac{AE}{EC} = 1$, that is, $AE = EC$, and $E$ is the mid-point of $AC$.

Question 8

Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Solution. Let $D$ and $E$ be the mid-points of $AB$ and $AC$ in $\triangle ABC$. This time the lengths are what we know and the parallel line is what we must prove, which is why the converse is the right tool.

Since $AD = DB$ and $AE = EC$,

$$\frac{AD}{DB} = 1 = \frac{AE}{EC}.$$

So $DE$ divides $AB$ and $AC$ in the same ratio, and by Theorem 6.2, $DE \parallel BC$.

Question 9

$ABCD$ is a trapezium in which $AB \parallel DC$ and its diagonals intersect each other at the point $O$. Show that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$.

Solution. The Basic Proportionality Theorem needs a line parallel to a side of a triangle, and the diagonals alone do not provide one. The hint printed with the textbook’s answers supplies it: through $O$, draw a line parallel to $DC$, meeting $AD$ at $E$ and $BC$ at $F$. Since $AB \parallel DC$, this line is parallel to $AB$ as well.

In $\triangle ADC$, $EO \parallel DC$, with $E$ on $AD$ and $O$ on $AC$, so

$$\frac{AE}{ED} = \frac{AO}{OC}. \qquad (1)$$

In $\triangle DAB$, $EO \parallel AB$, with $E$ on $DA$ and $O$ on $DB$, so

$$\frac{DE}{EA} = \frac{DO}{OB}, \quad\text{that is,}\quad \frac{AE}{ED} = \frac{BO}{OD}. \qquad (2)$$

From (1) and (2),

$$\frac{AO}{OC} = \frac{BO}{OD}.$$

Cross-multiplying gives $AO \cdot OD = BO \cdot OC$, and dividing both sides by $BO \cdot OD$ gives $\dfrac{AO}{BO} = \dfrac{CO}{DO}$.

Only the part $EO$ of the constructed line is needed. Working with $OF$ instead, in $\triangle BDC$ and $\triangle CAB$, gives the same equation.

Question 10

The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$. Show that $ABCD$ is a trapezium.

Solution. This is the converse of question 9, so the tool changes from Theorem 6.1 to Theorem 6.2. We have to show that one pair of opposite sides is parallel, and we may not assume it.

Through $O$ draw $OE \parallel AB$, meeting $AD$ at $E$. In $\triangle DAB$, by the Basic Proportionality Theorem,

$$\frac{AE}{ED} = \frac{BO}{OD}. \qquad (1)$$

The given condition $\dfrac{AO}{BO} = \dfrac{CO}{DO}$ cross-multiplies to $AO \cdot DO = BO \cdot CO$, and dividing both sides by $CO \cdot DO$ gives

$$\frac{AO}{OC} = \frac{BO}{OD}. \qquad (2)$$

From (1) and (2),

$$\frac{AE}{ED} = \frac{AO}{OC}.$$

Now look at $\triangle ADC$. The line $EO$ divides its sides $AD$ and $AC$ in the same ratio, so by Theorem 6.2, $EO \parallel DC$.

We now have $AB \parallel EO$ and $EO \parallel DC$, so $AB \parallel DC$. A quadrilateral with a pair of opposite sides parallel is a trapezium.

Common mistakes

• Question 1(ii), writing $\frac{AD}{AB} = \frac{AE}{EC}$. A part-to-whole ratio on one side and a part-to-part ratio on the other is the most common error with this theorem. Both sides must be the same type: $\frac{AD}{DB} = \frac{AE}{EC}$ or $\frac{AD}{AB} = \frac{AE}{AC}$. The $7.2$ cm in the figure is $DB$, not $AB$.
• Question 2(iii), comparing $\frac{PE}{PQ}$ with $\frac{PF}{FR}$. The data give the whole sides $PQ$ and $PR$, so either compare $\frac{PE}{PQ}$ with $\frac{PF}{PR}$, or subtract first to find $EQ$ and $FR$. Mixing the two forms produces a false “No”.
• Question 2(i), answering “No” without a reason. The ratios $1.3$ and $1.5$ differ, and it is Theorem 6.1 that turns this into a definite answer: if $EF$ were parallel to $QR$, the ratios would have to be equal.
• Questions 3 and 4, applying the theorem in the wrong triangle. The parallel line and the side it is parallel to must belong to the same triangle. In question 4, $DF \parallel AE$ is used in $\triangle ABE$, not $\triangle ABC$, because $AE$ is a side of $\triangle ABE$ only.
• Questions 5 and 6, stopping at equal ratios. Equal ratios say nothing about parallel lines until the converse is quoted, and it must be quoted for the triangle that contains both sides: $\triangle PQR$ in question 5, $\triangle OQR$ in question 6.
• Questions 7 and 8, swapping the theorems. Question 7 starts from a parallel line and concludes something about lengths, so it needs Theorem 6.1. Question 8 starts from lengths and concludes a parallel line, so it needs Theorem 6.2.
• Question 10, assuming the answer. It is tempting to write “since $AB \parallel DC$” as in question 9, but that is the conclusion here. The line through $O$ is drawn parallel to one side only, and its parallelism to the other side has to come from the converse.

Practise next

• Exercise 6.3 — the AA, SSS and SAS similarity criteria, which rest on this theorem; its question 3 proves the trapezium result of question 9 again with similar triangles and no construction.
• Exercise 6.1 — worth revisiting for the definition of similar polygons if the idea of proportional sides still feels loose.