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Q4:

In Fig. 6.36, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\triangle PQS \sim \triangle TQR$.

Solution :

Given:

1. In $\triangle PQR$, $\angle 1 = \angle 2$ (where $\angle 1 = \angle PQR$ and $\angle 2 = \angle PRQ$).

2. The ratio condition: $\frac{QR}{QS} = \frac{QT}{PR}$.

To Prove:

$\triangle PQS \sim \triangle TQR$

P Q R T S

Step 1: Analyze the properties of $\triangle PQR$

In $\triangle PQR$, we are given that $\angle 1 = \angle 2$.

Let $\angle 1 = \angle PQR$ and $\angle 2 = \angle PRQ$.

Since the angles opposite to equal sides are equal, and conversely, if two angles of a triangle are equal, the sides opposite to them are equal [By the Converse of Isosceles Triangle Theorem].

Therefore, $PQ = PR$.

Step 2: Modify the given ratio

The given equation is: $\frac{QR}{QS} = \frac{QT}{PR}$.

Substitute $PR = PQ$ (derived in Step 1) into the equation:

$\frac{QR}{QS} = \frac{QT}{PQ}$

Rearranging the terms to align with the sides of the triangles we intend to prove similar ($\triangle PQS$ and $\triangle TQR$):

$\frac{QS}{QR} = \frac{PQ}{QT}$

Step 3: Establish Similarity using SAS Criterion

Consider $\triangle PQS$ and $\triangle TQR$:

1. From Step 2, we have $\frac{QS}{QR} = \frac{PQ}{QT}$ (or equivalently $\frac{PQ}{QS} = \frac{QT}{QR}$).

2. Observe the common angle between these sides: $\angle PQS = \angle TQR$ (This is the same angle, $\angle 1$).

Since one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, the triangles are similar [By SAS (Side-Angle-Side) Similarity Criterion].

Conclusion:

Therefore, $\triangle PQS \sim \triangle TQR$.

Final Answer: Since the ratio of the sides including the common angle $\angle Q$ are proportional ($\frac{PQ}{QT} = \frac{QS}{QR}$), $\triangle PQS \sim \triangle TQR$ by the SAS similarity criterion.


More Questions from Class 10 Mathematics Triangles EXERCISE 6.3


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