Ex 6.6, 1 In Fig. 6.56, PS is the bisector of $\angle QPR$ of $\triangle PQR$. Prove that $\frac{QS}{SR} = \frac{PQ}{PR}$. Given: $\triangle PQR$ and PS is the bisector of $\angle QPR$ i.e. $\angle QPS = \angle RPS$ To Prove: $\frac{QS}{SR} = \frac{PQ}{PR}$

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Step-by-Step Solution

Step 1: Construction

To prove the Angle Bisector Theorem, we need to construct a parallel line. We draw a line through point $R$ parallel to $PS$, and extend $QP$ to meet this line at point $T$. This construction creates a transversal line $QR$ intersecting parallel lines $PS$ and $RT.

Step 2: Identify Equal Angles

Since $PS \parallel RT$ and $QT$ is a transversal, the corresponding angles $\angle QPS$ and $\angle QTR$ are equal. Also, since $PS \parallel RT$ and $PR$ is a transversal, the alternate interior angles $\angle SPR$ and $\angle PRT$ are equal. We are given that $PS$ is the angle bisector of $\angle QPR$, so $\angle QPS = \angle SPR.

Step 3: Equate Angles in Triangle PRT

From the previous step, we have $\angle QPS = \angle QTR$ and $\angle SPR = \angle PRT$. Since $\angle QPS = \angle SPR$ (given), it follows that $\angle QTR = \angle PRT.

Step 4: Identify Equal Sides in Triangle PRT

In $\triangle PRT$, since $\angle QTR = \angle PRT$, the sides opposite to these angles must be equal. Therefore, $PR = PT.

Step 5: Apply Basic Proportionality Theorem

Now consider $\triangle QRT$. We have constructed $PS \parallel RT$. By the Basic Proportionality Theorem (also known as Thales Theorem or Intercept Theorem), if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides proportionally. Therefore, $\frac{QS}{SR} = \frac{QP}{PT}.

Step 6: Substitute and Conclude

From Step 4, we established that $PT = PR$. Substituting $PR$ for $PT$ in the equation from Step 5, we get $\frac{QS}{SR} = \frac{PQ}{PR}$. This proves the Angle Bisector Theorem.