CONTENT
- The angle which an arc subtends at the centre is twice the angle it subtends at the circumference.
- Angles in the same segment of a circle are equal.
- Angle in a semi-circle.
- Tangent to a circle.
PROOF OF (i) The angle which an arc subtends at the centre is twice the angle it subtends at the circumference.
The angle which an arc (or a chord) of a circle subtends at the centre of the circle is twice the angle which it subtends at any point on the remaining part of the circumference.
Note: An arc of a circle is any connected part of the circle’s circumference.
A chord which is not a diameter divides the circle into two arcs- a major and a minor arc.
Given: An arc AB of a circle with ‘O’ and a point ‘P’ on the circumference.
To Prove: \(A\hat{O}B = 2A\hat{P}B\)
Construction: Join \(\overline{PO}\) and produce the line to a point D
Sketch:
Proof: Since \(\overline{AO} = \overline{OP}\) (radii in the same circle)
\(X_1 = X_2\) (base angles of isosceles \(A\hat{O}P\))
\(A\hat{O}D = X_1 + X_2\) (exterior angle of \(A\hat{O}P\))
\(A\hat{O}D = 2X_1\) (since \(X_1 = X_2\))
Similarly, \(B\hat{O}D = 2Y_1\)
In (i) acute/obtuse \(A\hat{O}B = A\hat{O}D + B\hat{O}D\)
In (ii) reflex \(A\hat{O}B = A\hat{O}D + B\hat{O}D \\ = 2X_1 + 2Y_1 \\ = 2(X_1 + Y_1) \\ = 2A\hat{P}B \)
In (iii) AOB = AOD – BOD
\(A\hat{O}B = A\hat{O}D -B\hat{O}D \\ = 2Y_1 -2X_1 \\ = 2(Y_1 -X_1) \\ = 2A\hat{P}B \\ \)
\( ∴ A\hat{O}B = 2A\hat{P}B\) (In all cases)
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