Inscribed Angles and Central Angles
An inscribed angle subtending the same arc is 1/2 of the central angle. All inscribed angles subtending the same arc are equal.
Cyclic Quadrilaterals
The sum of the opposite angles is 180° (∠A+∠C=∠B+∠D=180°)
Example ① (Inscribed Angle)
In inscribed △ABC, if the central angle subtending arc BC is 80°, find ∠BAC.
Solution:∠BAC = 80°/2 = 40°
In inscribed △ABC, if the central angle subtending arc BC is 80°, find ∠BAC.
Solution:∠BAC = 80°/2 = 40°
Power of a Point Theorem
Pattern ① (Intersection Inside the Circle): When two chords PQ and RS of a circle intersect at point T
TP × TQ = TR × TS
Pattern ② (Point Outside the Circle): When two secants drawn from an external point T intersect the circle at P, Q and R, S
TP × TQ = TR × TS
Pattern ③ (Tangent and Secant): When a tangent drawn from external point T touches the circle at P, and a secant intersects the circle at Q, R
TP² = TQ × TR
Example ② (Power of a Point Theorem: Internal Intersection)
Two chords AB and CD intersect at point T inside a circle. If TA=3, TB=4, and TC=2, find TD.
TA×TB=TC×TD → 3×4=2×TD → TD=6
Solution:TD = 6
Two chords AB and CD intersect at point T inside a circle. If TA=3, TB=4, and TC=2, find TD.
TA×TB=TC×TD → 3×4=2×TD → TD=6
Solution:TD = 6
Example ③ (Tangent and Secant)
From external point T, draw a tangent touching the circle at P. A secant from T intersects the circle at Q and R (TQ＜TR). If TQ=3 and TR=12, find TP.
TP²=TQ×TR=3×12=36 → TP=6
Solution:TP = 6
From external point T, draw a tangent touching the circle at P. A secant from T intersects the circle at Q and R (TQ＜TR). If TQ=3 and TR=12, find TP.
TP²=TQ×TR=3×12=36 → TP=6
Solution:TP = 6
Key Points
- Internal intersection: TP×TQ=TR×TS (intersection of two chords)
- External point: TP×TQ=TR×TS (two secants)
- Tangent: TP²=TQ×TR (tangent and secant)
- Cyclic quadrilateral: use the converse of the sum of opposite angles=180° to prove that “four points lie on the same circle”
Practice Problems
- Chords AB and CD intersect at point T inside a circle. If TA=5, TB=4, and TC=10, find TD.
- Two secants are drawn from external point T. On one, TQ=2 and TR=8; on the other, TS=4. Find TU (U is the farther intersection point).
- In cyclic quadrilateral ABCD, if ∠A=70°, find ∠C.
Solutions and Explanations
- 5×4=10×TD → TD=2
- TQ×TR=TS×TU → 2×8=4×TU → TU=4
- ∠A+∠C=180° → ∠C=180°−70°=110°