How to apply Basic Proportionality Theorem (Thales Theorem)?

If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio: PM/MQ = PN/NR.

How do diagonals of a trapezium divide each other?

In a trapezium with parallel sides KL and NM, drawing a line parallel to the bases through diagonal intersection O proves that diagonals divide proportionally: KO/LO = MO/NO.

Class 10 Maths Chapter 6 Exercise 6.2 NCERT Solutions

Questions 1 & 2 Basic Proportionality Theorem & Converse

1. In the figures below, $MN \parallel QR$. Find $NR$ in (i) and $PM$ in (ii):

Figure (i)

Figure (ii)

2. $X$ and $Y$ are points on sides $DE$ and $DF$ respectively of a $\Delta DEF$. State whether $XY \parallel EF$ in each case:
(i) $DX = 4\text{ cm}, XE = 5\text{ cm}, DY = 8\text{ cm}, YF = 10\text{ cm}$
(ii) $DX = 3\text{ cm}, XE = 4\text{ cm}, DY = 4\text{ cm}, YF = 5\text{ cm}$
(iii) $DE = 10\text{ cm}, DF = 15\text{ cm}, DX = 2\text{ cm}, DY = 3\text{ cm}$

Question 1 Solution

(i) In $\Delta PQR$, since $MN \parallel QR$, by Basic Proportionality Theorem (Thales Theorem):
$\dfrac{PM}{MQ} = \dfrac{PN}{NR}$
$\dfrac{2}{4} = \dfrac{3}{NR} \implies \dfrac{1}{2} = \dfrac{3}{NR} \implies NR = 3 \times 2 = \mathbf{6\text{ cm}}$.

(ii) In $\Delta PQR$, since $MN \parallel QR$, by BPT:
$\dfrac{PM}{MQ} = \dfrac{PN}{NR}$
$\dfrac{PM}{5} = \dfrac{2.4}{6} \implies \dfrac{PM}{5} = 0.4 \implies PM = 5 \times 0.4 = \mathbf{2\text{ cm}}$.

Question 2 Solution

By Converse of BPT, if a line divides two sides in the same ratio, it is parallel to the third side.

(i) $\dfrac{DX}{XE} = \dfrac{4}{5} = 0.8$; $\dfrac{DY}{YF} = \dfrac{8}{10} = 0.8$.
Since $\dfrac{DX}{XE} = \dfrac{DY}{YF} = 0.8$, $\therefore XY \parallel EF$ (Converse of BPT holds).

(ii) $\dfrac{DX}{XE} = \dfrac{3}{4} = 0.75$; $\dfrac{DY}{YF} = \dfrac{4}{5} = 0.8$.
Since $\dfrac{DX}{XE} \neq \dfrac{DY}{YF}$, $\therefore XY$ is NOT parallel to $EF$.

(iii) $XE = DE - DX = 10 - 2 = 8\text{ cm}$; $YF = DF - DY = 15 - 3 = 12\text{ cm}$.
$\dfrac{DX}{XE} = \dfrac{2}{8} = \dfrac{1}{4} = 0.25$; $\dfrac{DY}{YF} = \dfrac{3}{12} = \dfrac{1}{4} = 0.25$.
Since $\dfrac{DX}{XE} = \dfrac{DY}{YF}$, $\therefore XY \parallel EF$ (Converse of BPT holds).

Questions 3, 4, 5 & 6 Applying BPT Across Composite Figures

3. In the figure, if $XY \parallel BC$ and $XZ \parallel DC$, prove that $\dfrac{AY}{AB} = \dfrac{AZ}{AD}$.

4. In the figure, $SU \parallel PT$ and $ST \parallel PR$. Prove that $\dfrac{QU}{UT} = \dfrac{QT}{TR}$.

5. In the figure, $AB \parallel XP$ and $AC \parallel XQ$. Show that $BC \parallel PQ$.

6. In the figure, $A, B$ and $C$ are points on $OX, OY$ and $OZ$ respectively such that $AB \parallel XY$ and $AC \parallel XZ$. Show that $BC \parallel YZ$.

Question 3 Proof

In $\Delta ABC$, since $XY \parallel BC$, by BPT:
$\dfrac{AY}{YB} = \dfrac{AX}{XC} \implies \dfrac{AY}{AY + YB} = \dfrac{AX}{AX + XC} \implies \dfrac{AY}{AB} = \dfrac{AX}{AC}$ (1)

In $\Delta ADC$, since $XZ \parallel DC$, by BPT:
$\dfrac{AZ}{ZD} = \dfrac{AX}{XC} \implies \dfrac{AZ}{AD} = \dfrac{AX}{AC}$ (2)

From (1) and (2):
$\dfrac{AY}{AB} = \dfrac{AZ}{AD}$. Hence proved.

Question 4 Proof

In $\Delta PQT$, since $SU \parallel PT$, by BPT:
$\dfrac{QS}{SP} = \dfrac{QU}{UT}$ (1)

In $\Delta PQR$, since $ST \parallel PR$, by BPT:
$\dfrac{QS}{SP} = \dfrac{QT}{TR}$ (2)

Comparing (1) and (2):
$\dfrac{QU}{UT} = \dfrac{QT}{TR}$. Hence proved.

Question 5 Proof

In $\Delta XOP$, since $AB \parallel XP$, by BPT:
$\dfrac{OA}{AX} = \dfrac{OB}{BP}$ (1)

In $\Delta XOQ$, since $AC \parallel XQ$, by BPT:
$\dfrac{OA}{AX} = \dfrac{OC}{CQ}$ (2)

From (1) and (2): $\dfrac{OB}{BP} = \dfrac{OC}{CQ}$.
Now in $\Delta OPQ$, the line segment $BC$ divides sides $OP$ and $OQ$ in the same ratio.
$\therefore$ By the Converse of Basic Proportionality Theorem, $BC \parallel PQ$. Hence proved.

Question 6 Proof

In $\Delta OXY$, since $AB \parallel XY$, by BPT:
$\dfrac{OA}{AX} = \dfrac{OB}{BY}$ (1)

In $\Delta OXZ$, since $AC \parallel XZ$, by BPT:
$\dfrac{OA}{AX} = \dfrac{OC}{CZ}$ (2)

From (1) and (2): $\dfrac{OB}{BY} = \dfrac{OC}{CZ}$.
In $\Delta OYZ$, since $\dfrac{OB}{BY} = \dfrac{OC}{CZ}$, by the Converse of Basic Proportionality Theorem:
$BC \parallel YZ$. Hence proved.

Questions 7 & 8 Proving Mid-Point Theorems via Similarity

7. Using Theorem 6.1 (BPT), prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

8. Using Theorem 6.2 (Converse BPT), prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Question 7 Solution

Let $\Delta PQR$ be a triangle where $D$ is the mid-point of side $PQ$, and a line through $D$ parallel to $QR$ intersects $PR$ at $E$ ($DE \parallel QR$).
Since $D$ is the mid-point of $PQ$: $PD = DQ \implies \dfrac{PD}{DQ} = 1$ (1)
By BPT, since $DE \parallel QR$: $\dfrac{PD}{DQ} = \dfrac{PE}{ER}$ (2)
Substituting (1) into (2):
$1 = \dfrac{PE}{ER} \implies PE = ER$.
Since $PE = ER$, point $E$ bisects side $PR$. Hence proved.

Question 8 Solution

Let $\Delta ABC$ be a triangle where $X$ is the mid-point of $AB$ and $Y$ is the mid-point of $AC$.
Since $X$ is the mid-point of $AB$: $AX = XB \implies \dfrac{AX}{XB} = 1$ (1)
Since $Y$ is the mid-point of $AC$: $AY = YC \implies \dfrac{AY}{YC} = 1$ (2)
From (1) and (2): $\dfrac{AX}{XB} = \dfrac{AY}{YC}$.
Since line segment $XY$ divides sides $AB$ and $AC$ in the same ratio, by Theorem 6.2 (Converse of BPT):
$XY \parallel BC$. Hence proved.

Questions 9 & 10 Trapezium Diagonals & Ratio Properties

9. $KLMN$ is a trapezium with $KL \parallel NM$. Its diagonals intersect each other at $O$. Show that $\dfrac{KO}{LO} = \dfrac{MO}{NO}$.

10. The diagonals of a quadrilateral $PQRS$ intersect each other at $O$ such that $\dfrac{PO}{QO} = \dfrac{RO}{SO}$. Show that $PQRS$ is a trapezium.

Question 9 Proof

Given: Trapezium $KLMN$ with $KL \parallel NM$ and diagonals intersecting at $O$.
Construction: Through point $O$, draw line $OE \parallel KL$ meeting $KN$ at $E$. Since $KL \parallel NM$, $OE \parallel NM$.

In $\Delta KNL$, since $OE \parallel KL$, by BPT:
$\dfrac{NE}{EK} = \dfrac{NO}{OL}$ (1)

In $\Delta KNM$, since $OE \parallel NM$, by BPT:
$\dfrac{NE}{EK} = \dfrac{MO}{OK}$ (2)

Equating (1) and (2):
$\dfrac{NO}{OL} = \dfrac{MO}{OK} \implies \dfrac{OK}{OL} = \dfrac{MO}{NO} \implies \dfrac{KO}{LO} = \dfrac{MO}{NO}$. Hence proved.

Question 10 Proof

Given: Quadrilateral $PQRS$ with diagonals intersecting at $O$ such that $\dfrac{PO}{QO} = \dfrac{RO}{SO} \implies \dfrac{PO}{RO} = \dfrac{QO}{SO}$.
Construction: Draw line segment $OE \parallel PQ$ meeting $PS$ at $E$.

In $\Delta SPQ$, since $OE \parallel PQ$, by BPT:
$\dfrac{SE}{EP} = \dfrac{SO}{OQ} = \dfrac{SO}{QO}$ (1)

From the given relation, $\dfrac{QO}{SO} = \dfrac{PO}{RO} \implies \dfrac{SO}{QO} = \dfrac{RO}{PO}$.
Substituting into (1): $\dfrac{SE}{EP} = \dfrac{RO}{PO}$.

In $\Delta PSR$, since $\dfrac{SE}{EP} = \dfrac{SO}{PO}$ (i.e. line $OE$ divides sides $SP$ and $SR$ proportionally), by Converse of BPT:
$OE \parallel SR$.

Since $PQ \parallel OE$ and $OE \parallel SR$, it follows that $PQ \parallel SR$.
Since a pair of opposite sides is parallel, $PQRS$ is a trapezium. Hence proved.

Triangles - Exercise 6.2

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