A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP? Explain your reasoning.
Step-by-step solution
To find: The relation between ∠MOP and ∠MNP
Idea: Angles subtended by the same arc at points on the remaining part of the circle are equal (each is half the angle at the centre). Here both N and O see the arc MP that does not contain them.
- Going round the circle in the order M, N, O, P, the points N and O both lie on the arc from M to P that passes through N and O. So they are outside the other arc MP.½ mark
- ∠MNP and ∠MOP are both angles subtended by that other arc MP at points of the circle outside it. Each equals half the angle the arc subtends at the centre, so ∠MOP = ∠MNP (angles in the same segment).1½ marks
- Using the diameter: ∠MPN = 90° and ∠MON = 90° (angles in a semicircle). In right ΔMPN, ∠MNP = 90° − ∠PMN; so ∠MOP = ∠MNP = 90° − ∠PMN.1 mark
Answer to write in the exam
N and O lie on the same side of chord MP (vertices in order M, N, O, P).
∠MOP = ∠MNP (angles in the same segment; each = ½ angle of arc MP at the centre)
MN diameter ⇒ ∠MPN = 90° and ∠MON = 90° (angles in a semicircle)
∴ ∠MOP = ∠MNP = 90° − ∠PMN
Common mistakes that cost marks
- Taking O to be the centre of the circle. In this question O is a vertex of the quadrilateral MNOP.
- Saying the two angles add up to 180°. That is for opposite angles of the quadrilateral; ∠MOP and ∠MNP stand on the same chord from the same side.
- Forgetting the reason: write “angles in the same segment” (or “subtended by the same arc”).
How this can come in the exam
PQRS is a cyclic quadrilateral and ∠PQS = 40°. Then ∠PRS is
- 20°
- 40°
- 80°
- 140°
Show answer
(B) 40°
∠PQS and ∠PRS both stand on chord PS from the same side, so they are equal: 40°.
Try one yourself
In the quadrilateral MNOP above (MN a diameter), ∠NMP = 25°. Find ∠MNP and ∠MOP.
Show answer
∠MPN = 90° (angle in a semicircle), so ∠MNP = 180° − 90° − 25° = 65°, and ∠MOP = ∠MNP = 65°.
More questions like this
- Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., ∠CDE = ∠ABC, where E is a point on the extension of side CD).
- “There is no chord of a circle that is longer than its diameter.” How do you justify this statement?
- Let A be any point within a given circle with centre O. Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA.
- How would you use the following figure to justify the statement that the angle in a semicircle is 90°?
- In a circle, two chords CC’ and DD’ are drawn perpendicular to a diameter AB. Prove that the segment MM’ joining the midpoints of the chords CD and C’ D’ is perpendicular to AB.