Question : P, Q, R and S are respectively the mid-points of sides AB, BC, CD, DA of a rhombus ABCD. These points are joined to form a quadrilateral PQRS. Is the quadrilateral cyclic ?
Statement-I : \(\angle \text{ABC} = 120°\)
Statement-II : \(\angle \text{BCD} = 60°\)
Which one of the following is correct in respect of the above Question and the Statements ?
Rhombus Midpoints Quadrilateral Cyclic Property
The question asks whether the quadrilateral PQRS, formed by joining the midpoints (P, Q, R, S) of the sides (AB, BC, CD, DA) of a rhombus ABCD, is a cyclic quadrilateral.
To determine this, we analyze the properties of the figure PQRS based on the properties of the rhombus ABCD.
Geometric Analysis of PQRS
Joining the midpoints of the sides of any quadrilateral results in a parallelogram (a consequence of the midpoint theorem, often referred to as Varignon's theorem). Let's apply this:
- In \(\triangle \text{ABC}\), P and Q are midpoints of AB and BC. By the midpoint theorem, \(PQ \parallel \text{AC}\) and \(PQ = \frac{1}{2} \text{AC}\).
- In \(\triangle \text{ADC}\), S and R are midpoints of AD and CD. By the midpoint theorem, \(SR \parallel \text{AC}\) and \(SR = \frac{1}{2} \text{AC}\).
- From these properties, we see that \(PQ \parallel SR\) and \(PQ = SR\).
- Similarly, applying the midpoint theorem to \(\triangle \text{ABD}\) and \(\triangle \text{CBD}\), we get \(PS \parallel \text{BD}\) and \(QR \parallel \text{BD}\), with \(PS = QR = \frac{1}{2} \text{BD}\).
- Therefore, PQRS is a parallelogram.
Now, let's use the properties specific to a rhombus ABCD:
- The diagonals of a rhombus are perpendicular bisectors of each other. So, \(\text{AC} \perp \text{BD}\).
- Since \(PQ \parallel \text{AC}\) and \(PS \parallel \text{BD}\), the angle between PQ and PS must be equal to the angle between AC and BD.
- Therefore, the angle \(\angle \text{QPS}\) between adjacent sides PQ and PS must be \(90°\).
- A parallelogram with one angle measuring \(90°\) is a rectangle. Hence, PQRS is a rectangle.
Cyclic Nature of Rectangles
A rectangle is always a cyclic quadrilateral. This is because all its interior angles are \(90°\). The sum of opposite angles in PQRS is \(\angle \text{QPS} + \angle \text{QRS} = 90° + 90° = 180°\) and \(\angle \text{PQR} + \angle \text{PSR} = 90° + 90° = 180°\). A quadrilateral is cyclic if and only if the sum of its opposite angles is \(180°\).
Conclusion: The quadrilateral PQRS formed by joining the midpoints of the sides of any rhombus is always a rectangle, and therefore, it is always cyclic.
Evaluating the Statements
The fact that PQRS is cyclic is a general property derived from ABCD being a rhombus, independent of the specific angle values.
- Statement-I: \(\angle \text{ABC} = 120°\). This defines a specific rhombus, but the conclusion that PQRS is cyclic holds true regardless.
- Statement-II: \(\angle \text{BCD} = 60°\). This also defines a specific rhombus, but again, the conclusion remains the same.
Since the question "Is the quadrilateral cyclic?" can be answered affirmatively based on the properties of a rhombus alone, neither Statement I nor Statement II is needed individually or jointly to reach the conclusion.
Final Answer Determination
The question can be answered without reference to the specific angle information provided in the statements.
This aligns with Option D: The Question can be answered even without using both the Statements.
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