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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 $BC \parallel QR$.

 

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:

$BC \parallel QR$.

O P Q R A B C

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


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