Perpendicular from the Centre to a Chord
• A perpendicular from the centre to a chord always bisects the chord.
• If the centre joins the midpoint of a chord, the line is perpendicular to the chord.
• Chords equidistant from the centre are equal in length.
Central Angle and Circumference (Inscribed) Angle
• Central angle is made by two radii at the centre.
• Inscribed angle is made by two chords on the circle.
• Central angle equals its intercepted arc.
• Major arc > semicircle, minor arc < semicircle.
Relation Between Inscribed Angle and Its Arc
• Inscribed angle = ½ × central angle on the same arc.
• Therefore: central angle = 2 × inscribed angle.
• Angle in a semicircle (diameter as arc) = 90°.
Angles in the Circumference on the Same Arc
• All inscribed angles standing on the same arc are equal.
• Example: ∠ACB = ∠ADB if both stand on arc AB.
Cyclic Quadrilateral
• A quadrilateral whose all vertices lie on a circle.
• Opposite angles are supplementary (add to 180°).
• Exterior angle equals the interior opposite angle.
• All angles subtended by the same arc are equal.
Intersection of Chords Inside a Circle
• When two chords intersect, their opposite angles stand on equal arcs.
• Helps in finding unknown angles.
Angle Standing on the Diameter
• Any angle standing on a diameter is always a right angle (90°).
• Used to identify right triangles in circle-based questions.
Five Important Solved Questions
1. In a circle, ∠PQR = 100°. Find ∠OPR.
Solution:
Central angle POR = 2 × 100° = 200°.
Acute POR = 360 – 200 = 160°.
In isosceles ΔPOR: 160 + 2x = 180 → x = 10°.
∠OPR = 10°.
2. In triangle ABC, ∠ABC = 74°, ∠ACB = 30°. Find ∠BDC.
Solution:
∠BAC = 180 – (74 + 30) = 76°.
Angle on same arc BC at point D gives ∠BDC = 76°.
3. AC and BD intersect at E in cyclic quadrilateral ABCD. If ∠BEC = 130° and ∠ECD = 20°, find ∠BAC.
Solution:
∠CED = 180 – 130 = 50°.
In ΔCED: ∠EDC = 180 – (50 + 20) = 110°.
Angle on same arc BC: ∠BAC = 110°.
4. In a cyclic quadrilateral, AB is extended to E. Prove ∠ADC = ∠CBE.
Reason:
Opposite ∠ADC + ∠ABC = 180°.
Straight line: ∠ABC + ∠CBE = 180°.
Therefore: ∠ADC = ∠CBE.
5. In a semicircle, angle ∠ACB touches the circumference. Show it is 90°.
Solution:
Central angle on diameter = 180°.
Inscribed angle = ½ × 180° = 90°.
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