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Q6:
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 $BC \parallel QR$.
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 :
Given:
In $\triangle OPQ$, $A$ is on $OP$ and $B$ is on $OQ$ such that $AB \parallel PQ$.
In $\triangle OPR$, $A$ is on $OP$ and $C$ is on $OR$ such that $AC \parallel PR$.
To Prove:
$BC \parallel QR$.
Step 1: Applying Thales Theorem (Basic Proportionality Theorem) in $\triangle OPQ$
According to the Basic Proportionality Theorem (BPT), 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.
Since $AB \parallel PQ$ in $\triangle OPQ$:
$\frac{OA}{AP} = \frac{OB}{BQ}$ --- (Equation 1) [By BPT]
Step 2: Applying Thales Theorem in $\triangle OPR$
Since $AC \parallel PR$ in $\triangle OPR$:
$\frac{OA}{AP} = \frac{OC}{CR}$ --- (Equation 2) [By BPT]
Step 3: Comparing the Equations
From Equation 1 and Equation 2, we observe that the left-hand sides are identical ($\frac{OA}{AP}$). Therefore, we can equate the right-hand sides:
$\frac{OB}{BQ} = \frac{OC}{CR}$ --- (Equation 3)
Step 4: Applying the Converse of the Basic Proportionality Theorem
The Converse of the Basic Proportionality Theorem states that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
In $\triangle OQR$, we have established from Equation 3 that:
$\frac{OB}{BQ} = \frac{OC}{CR}$
Therefore, by the Converse of the Basic Proportionality Theorem, it follows that:
$BC \parallel QR$
Final Answer: Since the ratios of the segments on sides $OQ$ and $OR$ are equal ($\frac{OB}{BQ} = \frac{OC}{CR}$), by the Converse of the Basic Proportionality Theorem, $BC \parallel QR$ is proved.
More Questions from Class 10 Mathematics Triangles EXERCISE 6.2
- Q1(i): In Fig. 6.17, (i) and (ii), $DE \parallel BC$. Find $EC$ in (i).
- Q1(ii): In Fig. 6.17, (i) and (ii), $DE \parallel BC$. Find $AD$ in (ii).
- Q10: The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\frac{AO}{BO} = \frac{CO}{DO}$. Show that $ABCD$ is a trapezium.
- Q2(i): $E$ 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
- Q2(ii): $E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle PQR$. For each of the following cases, state whether $EF \parallel QR$ : (ii) $PE = 4$ cm, $QE = 4.5$ cm, $PF = 8$ cm and $RF = 9$ cm
- Q2(iii): $E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle PQR$. For each of the following cases, state whether $EF \parallel QR$ : (iii) $PQ = 1.28$ cm, $PR = 2.56$ cm, $PE = 0.18$ cm and $PF = 0.36$ cm
- Q3: In Fig. 6.18, if $LM \parallel CB$ and $LN \parallel CD$, prove that $\frac{AM}{AB} = \frac{AN}{AD}$.
- Q4: In Fig. 6.19, $DE \parallel AC$ and $DF \parallel AE$. Prove that $\frac{BF}{FE} = \frac{BE}{EC}$.
- Q5: In Fig. 6.20, $DE \parallel OQ$ and $DF \parallel OR$. Show that $EF \parallel QR$.
- Q7: 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).
- Q8: 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).
- Q9: $ABCD$ is a trapezium in which $AB \parallel DC$ and its diagonals intersect each other at the point $O$. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.
CBSE Solutions for Class 10 Mathematics Triangles
Chapters in CBSE - Class 10 Mathematics
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