NCERT Class 10 Exercise 6.2 Solutions: BPT and Its Converse

Solu­tions to Exer­cise 6.2#

Ques­tion 1#

In Fig. 6.17 (i) and (ii), $DE \parallel BC$. Find $EC$ in (i) and $AD$ in (ii). In the text­book fig­ure, part (i) has $AD = 1.5$ cm, $DB = 3$ cm and $AE = 1$ cm; part (ii) has $DB = 7.2$ cm, $AE = 1.8$ cm and $EC = 5.4$ cm.

(i) Since $DE \parallel BC$, The­o­rem 6.1 gives

$$\displaystyle \begin{aligned} \frac{AD}{DB} &= \frac{AE}{EC} \\ \frac{1.5}{3} &= \frac{1}{EC} \\ EC &= \frac{3 \times 1}{1.5} = 2 \end{aligned}$$

Check: $\displaystyle \frac{1.5}{3} = \frac{1}{2}$ and $\displaystyle \frac{1}{2} = \frac{1}{2}$.

Answer: $EC = 2$ cm.

(ii) Again by The­o­rem 6.1,

$$\displaystyle \begin{aligned} \frac{AD}{DB} &= \frac{AE}{EC} \\ \frac{AD}{7.2} &= \frac{1.8}{5.4} = \frac{1}{3} \\ AD &= \frac{7.2}{3} = 2.4 \end{aligned}$$

Check: $\displaystyle \frac{2.4}{7.2} = \frac{1}{3}$ and $\displaystyle \frac{1.8}{5.4} = \frac{1}{3}$.

Answer: $AD = 2.4$ 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 $EF \parallel QR$.

We use The­o­rem 6.2: $EF \parallel QR$ exactly when $\displaystyle \frac{PE}{EQ} = \frac{PF}{FR}$. Each case is also cross-checked by com­par­ing $PE \times FR$ with $PF \times EQ$.

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

$$\displaystyle \frac{PE}{EQ} = \frac{3.9}{3} = 1.3, \qquad \frac{PF}{FR} = \frac{3.6}{2.4} = 1.5.$$

The ratios are not equal. Cross-check: $3.9 \times 2.4 = 9.36$ while $3.6 \times 3 = 10.8$.

Answer: $EF$ is not par­al­lel to $QR$.

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

$$\displaystyle \frac{PE}{QE} = \frac{4}{4.5} = \frac{8}{9}, \qquad \frac{PF}{RF} = \frac{8}{9}.$$

The ratios are equal. Cross-check: $4 \times 9 = 36$ and $8 \times 4.5 = 36$.

Answer: $EF \parallel QR$.

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

The whole sides are given, so first find the remain­ing pieces:

$$EQ = PQ - PE = 1.28 - 0.18 = 1.10 \text{ cm}, \qquad FR = PR - PF = 2.56 - 0.36 = 2.20 \text{ cm}.$$

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

The ratios are equal. Cross-check: $0.18 \times 2.20 = 0.396$ and $0.36 \times 1.10 = 0.396$. (Quicker: $\displaystyle \frac{PE}{PQ} = \frac{0.18}{1.28} = \frac{9}{64}$ and $\displaystyle \frac{PF}{PR} = \frac{0.36}{2.56} = \frac{9}{64}$.)

Answer: $EF \parallel QR$.

Ques­tion 3#

In Fig. 6.18, if $LM \parallel CB$ and $LN \parallel CD$, prove that $\displaystyle \frac{AM}{AB} = \frac{AN}{AD}$. In the fig­ure, $ABCD$ is a quadri­lat­eral with diag­o­nal $AC$; $L$ lies on $AC$, $M$ on $AB$ and $N$ on $AD$.

Solu­tion.

In $\triangle ABC$: $M$ is on $AB$, $L$ is on $AC$ and $LM \parallel CB$. By The­o­rem 6.1 in whole-side form,

$$\displaystyle \frac{AM}{AB} = \frac{AL}{AC}. \qquad (1)$$

In $\triangle ADC$: $N$ is on $AD$, $L$ is on $AC$ and $LN \parallel CD$. By the same the­o­rem,

$$\displaystyle \frac{AN}{AD} = \frac{AL}{AC}. \qquad (2)$$

The right-hand sides of (1) and (2) are equal, so

$$\displaystyle \frac{AM}{AB} = \frac{AN}{AD}. \qquad \blacksquare$$

Why this works: the diag­o­nal $AC$ belongs to both tri­an­gles, so $\displaystyle \frac{AL}{AC}$ links them.

Ques­tion 4#

In Fig. 6.19, $DE \parallel AC$ and $DF \parallel AE$. Prove that $\displaystyle \frac{BF}{FE} = \frac{BE}{EC}$. In the fig­ure, $D$ is on $AB$, $E$ is on $BC$, and $F$ is on $BE$.

Solu­tion.

In $\triangle ABC$: $D$ is on $BA$, $E$ is on $BC$ and $DE \parallel AC$. By The­o­rem 6.1 (ratios from ver­tex $B$),

$$\displaystyle \frac{BD}{DA} = \frac{BE}{EC}. \qquad (1)$$

In $\triangle ABE$: $D$ is on $BA$, $F$ is on $BE$ and $DF \parallel AE$. By The­o­rem 6.1,

$$\displaystyle \frac{BD}{DA} = \frac{BF}{FE}. \qquad (2)$$

From (1) and (2),

$$\displaystyle \frac{BF}{FE} = \frac{BE}{EC}. \qquad \blacksquare$$

Check with the drawn case: $BD : DA = 2 : 1$, so both $\displaystyle \frac{BF}{FE}$ and $\displaystyle \frac{BE}{EC}$ equal $2$.

Ques­tion 5#

In Fig. 6.20, $DE \parallel OQ$ and $DF \parallel OR$. Show that $EF \parallel QR$. In the fig­ure, a point $O$ is joined to $P$, $Q$ and $R$; $D$ is on $PO$, $E$ on $PQ$ and $F$ on $PR$.

Solu­tion.

In $\triangle POQ$: $D$ is on $PO$, $E$ is on $PQ$ and $DE \parallel OQ$. By The­o­rem 6.1,

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

In $\triangle POR$: $D$ is on $PO$, $F$ is on $PR$ and $DF \parallel OR$. By The­o­rem 6.1,

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

From (1) and (2), $\displaystyle \frac{PE}{EQ} = \frac{PF}{FR}$. So in $\triangle PQR$, $E$ and $F$ divide $PQ$ and $PR$ in the same ratio. By The­o­rem 6.2,

$$EF \parallel QR. \qquad \blacksquare$$

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 $BC \parallel QR$.

Solu­tion.

In $\triangle OPQ$: $A$ is on $OP$, $B$ is on $OQ$ and $AB \parallel PQ$. By The­o­rem 6.1,

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

In $\triangle OPR$: $A$ is on $OP$, $C$ is on $OR$ and $AC \parallel PR$. By The­o­rem 6.1,

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

From (1) and (2), $\displaystyle \frac{OB}{BQ} = \frac{OC}{CR}$. So in $\triangle OQR$, $B$ and $C$ divide $OQ$ and $OR$ in the same ratio. By The­o­rem 6.2,

$$BC \parallel QR. \qquad \blacksquare$$

Why this works: this is Ques­tion 5 with the roles of the points renamed; the shared side $OP$ car­ries the link­ing ratio.

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.

Given: $\triangle ABC$, $D$ is the mid-point of $AB$, and the line through $D$ par­al­lel to $BC$ meets $AC$ at $E$.

To prove: $AE = EC$.

Proof. Since $D$ is the mid-point of $AB$, $AD = DB$, so $\displaystyle \frac{AD}{DB} = 1$. Since $DE \parallel BC$, The­o­rem 6.1 gives

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

So $AE = EC$, that is, $E$ is the mid-point of $AC$ and the line bisects the third side. $\blacksquare$

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.

Given: $\triangle ABC$ with $D$ the mid-point of $AB$ and $E$ the mid-point of $AC$.

To prove: $DE \parallel BC$.

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

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

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

Ques­tion 9#

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

Solu­tion.

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

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

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

In $\triangle DAB$: $E$ is on $DA$, $O$ is on $DB$ and $EO \parallel AB$. By The­o­rem 6.1 (ratios from ver­tex $D$),

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

From (1) and (2),

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

Cross-mul­ti­ply­ing, $AO \times OD = BO \times OC$. Divid­ing both sides by $BO \times OD$,

$$\displaystyle \frac{AO}{BO} = \frac{OC}{OD} = \frac{CO}{DO}. \qquad \blacksquare$$

Check with the drawn case, where $AB = 4$ and $DC = 8$: $\displaystyle \frac{AO}{OC} = \frac{BO}{OD} = \frac{1}{2}$, and indeed $\displaystyle \frac{AO}{BO} = \frac{CO}{DO}$.

Ques­tion 10#

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

Solu­tion. This is the con­verse of Ques­tion 9, so we aim for a pair of par­al­lel sides.

From $\displaystyle \frac{AO}{BO} = \frac{CO}{DO}$, cross-mul­ti­ply­ing gives $AO \times DO = BO \times CO$. Divid­ing both sides by $CO \times DO$,

$$\displaystyle \frac{AO}{CO} = \frac{BO}{DO}. \qquad (1)$$

Con­struc­tion: Through $O$ draw $EO \parallel AB$, meet­ing $AD$ at $E$.

In $\triangle DAB$: $E$ is on $DA$, $O$ is on $DB$ and $EO \parallel AB$. By The­o­rem 6.1,

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

From (1) and (2),

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

In $\triangle ADC$, $E$ is on $AD$ and $O$ is on $AC$, and they divide these sides in the same ratio. By The­o­rem 6.2, $EO \parallel DC$.

Now $EO \parallel AB$ (con­struc­tion) and $EO \parallel DC$. Lines par­al­lel to the same line are par­al­lel to each other, so $AB \parallel DC$.

A quadri­lat­eral with a pair of par­al­lel sides is a trapez­ium.

Answer: $AB \parallel DC$, so $ABCD$ is a trapez­ium. $\blacksquare$

Why this works: Ques­tions 9 and 10 use the same con­struc­tion. Ques­tion 9 goes from par­al­lel sides to equal ratios (The­o­rem 6.1); Ques­tion 10 goes from equal ratios back to par­al­lel sides (The­o­rem 6.2).

Key terms#

Basic Pro­por­tion­al­ity The­o­rem (The­o­rem 6.1)

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

Con­verse of BPT (The­o­rem 6.2)

A line divid­ing two sides of a tri­an­gle in the same ratio is par­al­lel to the third side.

Whole-side form

The con­se­quence $\displaystyle \frac{AD}{AB} = \frac{AE}{AC}$, which com­pares a piece with the whole side.

Mid-point the­o­rem

The seg­ment join­ing the mid-points of two sides of a tri­an­gle is par­al­lel to the third side; Ques­tions 7 and 8 prove it and its con­verse from The­o­rems 6.1 and 6.2.

Trapez­ium

A quadri­lat­eral with a pair of par­al­lel sides.

Con­struc­tion

An extra line drawn in a proof, such as $EO \parallel AB$ in Ques­tions 9 and 10.

Cross-mul­ti­pli­ca­tion

Replac­ing $\displaystyle \frac{a}{b} = \frac{c}{d}$ by $ad = bc$, use­ful for check­ing ratios with­out dec­i­mals.

Com­mon ques­tions#

How do I know which tri­an­gle to use in Ques­tions 3 to 6?#

Look at each par­al­lel pair. The tri­an­gle you need has the par­al­lel line inside it and the other line as one of its sides. For exam­ple, $DE \parallel AC$ in Ques­tion 4 points to $\triangle ABC$.

Why is a con­struc­tion needed in Ques­tions 9 and 10?#

The trapez­ium alone con­tains no tri­an­gle with a line par­al­lel to one of its sides through $O$. Draw­ing $EO \parallel AB$ cre­ates two such tri­an­gles.

In Ques­tion 2 (iii), can I com­pare $\displaystyle \frac{PE}{PQ}$ and $\displaystyle \frac{PF}{PR}$ directly?#

Yes. Those ratios are equal exactly when $\displaystyle \frac{PE}{EQ} = \frac{PF}{FR}$, and here both equal $\displaystyle \frac{9}{64}$.

Which the­o­rem do I quote when prov­ing lines par­al­lel?#

The­o­rem 6.2, the con­verse. The­o­rem 6.1 is quoted when the par­al­lel lines are given and a ratio is required.

Is every quadri­lat­eral with $\displaystyle \frac{AO}{BO} = \frac{CO}{DO}$ a trapez­ium?#

Yes, that is what Ques­tion 10 proves. A par­al­lel­o­gram also sat­is­fies the con­di­tion, since its diag­o­nals bisect each other; a par­al­lel­o­gram has a pair of par­al­lel sides.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.