Let P be an interior point of a triangle ABC , Let Q and R be the reflections of P in AB and AC , respectively if Q .A , R are collinear then `angle A ` equals -

Options

A. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `120^(@)`

Answer

To solve the problem, we need to analyze the given geometric configuration involving triangle \(ABC\) and the interior point \(P\). Here’s a step-by-step solution:

### Step 1: Understand the Configuration
- We have triangle \(ABC\) with point \(P\) inside it.
- Points \(Q\) and \(R\) are the reflections of point \(P\) across sides \(AB\) and \(AC\), respectively.

### Step 2: Draw the Diagram
- Draw triangle \(ABC\).
- Mark point \(P\) inside the triangle.
- Reflect point \(P\) across line \(AB\) to get point \(Q\).
- Reflect point \(P\) across line \(AC\) to get point \(R\).

### Step 3: Analyze the Collinearity Condition
- The problem states that points \(Q\), \(A\), and \(R\) are collinear.
- For three points to be collinear, the angles formed at point \(A\) must satisfy certain conditions.

### Step 4: Use Angle Relationships
- Let \( \angle PAB = \theta \) and \( \angle PAC = \phi \).
- When \(P\) is reflected to \(Q\) and \(R\), the angles change:
- The angle \( \angle QAB = \theta \)
- The angle \( \angle RAC = \phi \)

### Step 5: Set Up the Equation
- Since \(Q\), \(A\), and \(R\) are collinear, the angles around point \(A\) must sum to \(180^\circ\):
\[ \angle QAP + \angle RAP = 180^\circ \]
- This can be expressed as:
\[ \theta + \phi + \theta + \phi = 180^\circ \]
Simplifying gives:
\[ 2\theta + 2\phi = 180^\circ \]
Dividing by 2:
\[ \theta + \phi = 90^\circ \]

### Step 6: Conclusion
- The angle \(A\) in triangle \(ABC\) is given by:
\[ \angle A = \theta + \phi = 90^\circ \]
- Therefore, the measure of angle \(A\) is \(90^\circ\).

### Final Answer
\[ \angle A = 90^\circ \]

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