Basic Properties Of Circles Examples I

ExamplesIn the figure, AEC and BED are straight lines. It is given that \(\angle BDC = 30^\circ \), \(\angle ACB = 50^\circ \) and \(\angle ABC = 100^\circ \).
Prove that A, B, C and D are concyclic.

SolutionsIn ΔABC,
\(\begin{array}{1}\angle BAC + \angle ABC + \angle ACB = 180^\circ \\\angle BAC + 100^\circ + 50^\circ = 180^\circ \\\angle BAC = 30^\circ \end{array}\) (∠ sum of Δ)
∴ \(\angle BAC = \angle BDC = 30^\circ \)
∴ A, B, C and D are concyclic. (converse of ∠s in the same segment)

ExamplesIn the figure, O is the centre of the circle. AB is a diameter. P is an external point of the circle such that \(PA = PB\). PB intersects the circumference at C. AC intersects OP at D.
(a) Prove that PO ⊥ AB.
(b) Prove that O, A, P and C are concyclic.
(c) Prove ∠OPB = ∠OCA. 
Solutions(a) In ΔAPO and ΔBPO,
\(AP = BP\) (given)
\(AO = BO\) (radii)
\(OP = OP\) (common side)
∴ \(\Delta APO \cong \Delta BPO\) (S.S.S.)
∴ \(\angle POA = \angle POB\) (corr. ∠s, ≅ Δs)
\(\begin{array}{1}\angle POA + \angle POB = 180^\circ \\2\angle POA = 180^\circ \\\angle POA = 90^\circ \end{array}\) (adj. ∠s on st. line)
∴ \(PO \bot AB\)
(b) \(\angle ACB = 90^\circ \) (∠ in semicircle)
\(\begin{array}{1}\angle ACP = 180^\circ  90^\circ \\ = 90{^\circ }\end{array}\) (adj. ∠s on st. line)
∴ \(\angle AOP = \angle ACP = 90^\circ \)
∴ O, A, P and C are concyclic. (converse of ∠s in the same segment)
(c) \(\angle OPB = \angle OAC\) (∠s in the same segment)
∵ \(OA = OC\) (radii)
∴ \(\angle OAC = \angle OCA\) (base ∠s, isos. Δ)
∴ \(\angle OPB = \angle OCA\)

ExamplesIn the figure, O is the centre of two concentric circles. AB is the tangent to the smaller circle at M. Radii OA and OB of the larger circle intersect the circumference of the smaller circle at N and P respectively. It is given that \(OM = 20\) and \(AB = 30\).
(a) Find the length of OA.
(b) Find the length of AN. 
Solutions(a) ∵ \(\angle OMA = 90^\circ \) (tangent ⊥ radius)
∴ \(AM = BM\) (⊥ from centre bisects chord)
∴ \(\begin{array}{1}AM = \frac{1}{2}AB\\ = \frac{1}{2} \times 30\\ = 15\end{array}\)
In ΔOAM,
∵ \(O{A^2} = O{M^2} + A{M^2}\) (Pyth. theorem)
∴ \(\begin{array}{1}OA = \sqrt {O{M^2} + A{M^2}} \\ = \sqrt {{{20}^2} + {{15}^2}} \\ = \underline{\underline {25}} \end{array}\)
\(\begin{array}{1}(b)ON = OM\\ = 20\end{array}\) (radii)
\(\begin{array}{1}AN = OA  ON\\ = 25  20\\ = \underline{\underline 5} \end{array}\)