ABCD is a cyclic quadrilateral. The bisectors of the angles A, B, C and D cut the circle at P, Q, R and S respectively. What is ∠PQR + ∠RSP equal to?
180°
Understanding the Problem: Cyclic Quadrilaterals and Angle Bisectors
The problem describes a cyclic quadrilateral ABCD, which means all its vertices lie on a circle. The bisectors of each interior angle (∠A, ∠B, ∠C, ∠D) are drawn, and these bisectors intersect the circle at points P, Q, R, and S, respectively. We need to find the sum of two specific angles in the quadrilateral formed by these new points: ∠PQR + ∠RSP.
Since the points P, Q, R, and S all lie on the circle, the quadrilateral PQRS is also a cyclic quadrilateral.
Key Geometric Properties
We will use the following properties to solve this problem:
- Cyclic Quadrilateral Angle Property: The sum of opposite angles in a cyclic quadrilateral is 180°.
- Angle Bisector Property (in a Circle): An angle ∠XAY inscribed in a circle intercepts arc XY. If AZ is the bisector of ∠XAY, where Z is on the circle, then the arc XZ is equal in measure to the arc ZY. That is, \(m(\text{arc XZ}) = m(\text{arc ZY})\). This means Z is the midpoint of the arc XY that is intercepted by ∠XAY.
- Arcs and Angles: The measure of an inscribed angle is half the measure of its intercepted arc.
- Midpoint of an Arc: The line segment joining the center of the circle to the midpoint of a chord is perpendicular to the chord and bisects the arc subtended by the chord. The line containing the perpendicular bisector of a chord passes through the center of the circle.
Analyzing the Effect of Angle Bisectors
Let the circle be \(\mathcal{C}\). ABCD is a cyclic quadrilateral on \(\mathcal{C}\).
- The bisector of ∠A (ray AP) intersects the circle at P. ∠BAP = ∠DAP = ∠A/2. These inscribed angles intercept arcs BP and DP respectively. Therefore, \(m(\text{arc BP}) = m(\text{arc DP})\). This means P is the midpoint of the arc BD that does not contain A.
- The bisector of ∠B (ray BQ) intersects the circle at Q. ∠ABQ = ∠CBQ = ∠B/2. These inscribed angles intercept arcs AQ and CQ respectively. Therefore, \(m(\text{arc AQ}) = m(\text{arc CQ})\). This means Q is the midpoint of the arc AC that does not contain B.
- The bisector of ∠C (ray CR) intersects the circle at R. ∠BCR = ∠DCR = ∠C/2. These inscribed angles intercept arcs BR and DR respectively. Therefore, \(m(\text{arc BR}) = m(\text{arc DR})\). This means R is the midpoint of the arc BD that does not contain C.
- The bisector of ∠D (ray DS) intersects the circle at S. ∠CDS = ∠ADS = ∠D/2. These inscribed angles intercept arcs CS and AS respectively. Therefore, \(m(\text{arc CS}) = m(\text{arc AS})\). This means S is the midpoint of the arc AC that does not contain D.
Locating Points P, Q, R, S
Let's consider the arcs BD. There are two arcs BD. Let's call the arc BD that contains C as arc BD\(_1\) and the arc BD that contains A as arc BD\(_2\). Similarly, let arc AC\(_1\) contain D and arc AC\(_2\) contain B.
- P is the midpoint of arc BD\(_1\) (\(m(\text{arc BP}) = m(\text{arc DP})\)).
- R is the midpoint of arc BD\(_2\) (\(m(\text{arc BR}) = m(\text{arc DR})\)).
- Q is the midpoint of arc AC\(_1\) (\(m(\text{arc AQ}) = m(\text{arc CQ})\)).
- S is the midpoint of arc AC\(_2\) (\(m(\text{arc AS}) = m(\text{arc CS})\)).
The points P and R are midpoints of the two arcs formed by chord BD. This means the line segment PR passes through the center of the circle. Hence, PR is a diameter of the circle.
Similarly, the points Q and S are midpoints of the two arcs formed by chord AC. This means the line segment QS passes through the center of the circle. Hence, QS is a diameter of the circle.
Determining the Type of Quadrilateral PQRS
The quadrilateral PQRS is a cyclic quadrilateral because all its vertices lie on the circle. Its diagonals are PR and QS. We have established that both PR and QS are diameters of the circle.
A cyclic quadrilateral whose diagonals are diameters is a rectangle.
Therefore, the quadrilateral PQRS is a rectangle.
Calculating the Sum of Angles
In a rectangle, all interior angles are right angles, measuring 90°.
- ∠PQR = 90°
- ∠QRS = 90°
- ∠RSP = 90°
- ∠SPQ = 90°
The question asks for the sum ∠PQR + ∠RSP.
\(∠PQR + ∠RSP = 90° + 90° = 180°\).
Conclusion
The sum of the angles ∠PQR and ∠RSP is 180° because PQRS is a rectangle formed by the midpoints of the arcs subtended by the diagonals of the original cyclic quadrilateral ABCD.
| Step | Description | Result / Property |
|---|---|---|
| 1 | ABCD is a cyclic quadrilateral. | Vertices A, B, C, D are on the circle. |
| 2 | Angle bisectors of A, B, C, D meet the circle at P, Q, R, S. | P, Q, R, S are on the circle. PQRS is cyclic. |
| 3 | Angle bisector of A (AP) \(\implies m(\text{arc BP}) = m(\text{arc DP})\). | P is the midpoint of arc BD. |
| 4 | Angle bisector of B (BQ) \(\implies m(\text{arc AQ}) = m(\text{arc CQ})\). | Q is the midpoint of arc AC. |
| 5 | Angle bisector of C (CR) \(\implies m(\text{arc BR}) = m(\text{arc DR})\). | R is the midpoint of arc BD. |
| 6 | Angle bisector of D (DS) \(\implies m(\text{arc AS}) = m(\text{arc CS})\). | S is the midpoint of arc AC. |
| 7 | P and R are midpoints of the two arcs BD. | PR is a diameter. |
| 8 | Q and S are midpoints of the two arcs AC. | QS is a diameter. |
| 9 | PQRS is a cyclic quadrilateral with diameters as diagonals. | PQRS is a rectangle. |
| 10 | Angles of a rectangle. | ∠PQR = 90°, ∠RSP = 90°. |
| 11 | Sum of angles. | ∠PQR + ∠RSP = 90° + 90° = 180°. |
Revision Table: Cyclic Quadrilateral Geometry
| Concept | Description | Relevance Here |
|---|---|---|
| Cyclic Quadrilateral | A quadrilateral whose vertices lie on a circle. | ABCD and PQRS are cyclic quadrilaterals. |
| Inscribed Angle | An angle formed by two chords in a circle that have a common endpoint on the circle. | Angle bisectors form inscribed angles that intercept arcs. ∠BAP intercepts arc BP. |
| Arc Measure | The measure of the central angle that subtends the arc. | Equal inscribed angles subtend equal arcs (\(m(\text{arc BP}) = m(\text{arc DP})\)). |
| Midpoint of an Arc | A point that divides an arc into two equal arcs. | P, Q, R, S are midpoints of specific arcs determined by the original quadrilateral's diagonals. |
| Diameter | A chord passing through the center of the circle. | Diagonals PR and QS are diameters. |
| Rectangle | A quadrilateral with four right angles. A cyclic quadrilateral with perpendicular diagonals is a rectangle. A cyclic quadrilateral with diameters as diagonals is a rectangle. | PQRS is a rectangle. |
Additional Information: Properties of Angle Bisectors in Cyclic Quadrilaterals
The points P, Q, R, and S formed by the intersection of the angle bisectors of a cyclic quadrilateral ABCD with the circumcircle have some interesting properties:
- The quadrilateral PQRS is always a rectangle, regardless of the specific shape of the cyclic quadrilateral ABCD. This is because its diagonals PR and QS are diameters of the circumcircle.
- The sides of the rectangle PQRS are parallel to the diagonals of the original quadrilateral ABCD. Specifically, PQ is parallel to AC, QR is parallel to BD, RS is parallel to AC, and SP is parallel to BD.
- The diagonals of the rectangle PQRS (which are diameters of the circle) are perpendicular bisectors of the diagonals of the original cyclic quadrilateral ABCD. PR is the perpendicular bisector of BD, and QS is the perpendicular bisector of AC.
These properties illustrate a fascinating relationship between a cyclic quadrilateral and the figure formed by the points where its angle bisectors meet the circumcircle.
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