Basic properties of circles  (10.2)
Understand the angle properties of a circle

Theory
the angle subtended by an arc of a circle at the centre
is double the angle subtended by the arc at any point
on the remaining part of the circumference

ExamplesIn the figure, O is the centre of the circle. AB and AC are equal
chords. If \(\angle BAC = 42^\circ \), find \(\angle\)OBC and \(\angle\)ABO. 
Solutions\(\begin{array}{1} \angle BOC = 2 \times 42^\circ \\ = {84^\circ {}}\end{array}\) ( at centre =2 at ⊙ce)
In OBC,
\(OC = OB\) (radii)
\(\angle OCB = \angle OBC\) (base s, isos. )
\(\begin{array}{1} \angle OBC + \angle OCB + \angle BOC = 180^\circ \\2\angle OBC + 84^\circ = 180{^\circ {}}\\2\angle OBC = 96{^\circ {}}\\\angle OBC = {\underline{\underline {48^\circ }} {}}\end{array}\) ( sum of )
In ABC,
\(AC = AB\) (given)
\(\angle ACB = \angle ABC\) (base s, isos. )
\(\begin{array}{1} \angle ABC + \angle ACB + \angle BAC = 180^\circ \\2\angle ABC + 42^\circ = 180{^\circ {}}\\2\angle ABC = 138{^\circ {}}\\\angle ABC = 69{^\circ {}}\end{array}\) ( sum of )
\(\begin{array}{1} \angle ABO = 69^\circ  48^\circ \\ = {\underline{\underline {21^\circ }} ^{}}\end{array}\)

Theory
Two angles subtended by the same arc of a circle
at any part of the circumference are equals

ExamplesIn the figure, chords AC and BD intersect at E. Find \(w + x + y + z\).

Solutions\(\begin{array}{1}\angle ACD = \angle ABD\\ = {x^{}}\end{array}\) (s in the same segment)
\(\begin{array}{1}\angle ADB = \angle ACB\\ = {y^{}}\end{array}\) (s in the same segment)
In ACD,
\(\begin{array}{1}\angle DAC + \angle ADC + \angle DCA = 180^\circ \\w + (y + z) + x = {180^\circ{}}\end{array}\) ( sum of )
\(\therefore w + x + y + z = 180^\circ \)

Theory
the arcs are proportional to their corresponding angles at the circumference 
ExamplesIn the figure, chords AC and BD intersect at E. If \(A\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over D} = B\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over C} \),
prove that \(\Delta ADC \cong \Delta BCD\). 
SolutionsIn ADC and BCD,
\(\begin{array}{c}\frac{{\angle ACD}}{{\angle BDC}} = \frac{{A\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over D} }}{{B\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over C} }}\\ = {1^{}}\end{array}\) (arc and at ⊙ce in prop.)
\(\angle ACD = \angle BDC\)
\(\angle DAC = \angle CBD\) (s in the same segment)
\(DC = CD\) (common side)
\(\therefore \Delta ADC \cong \Delta BCD\) (A.A.S.)

Graph

Theorythe angle in a semicircle is a right angle

ExamplesIn the figure, O is the centre of the circle, AD and BE are diameters.
Chord AC intersects BE at G and chord CE intersects AD at F. CD // BE
and BC // AD. Prove that \(\Delta BCE \cong \Delta DCA\). 
SolutionsIn BCE and DCA,
\(BE = DA\) (diameters)
\(\angle BCE = \angle DCA = 90^\circ \) ( in semicircle)
\(\angle CBG = \angle AOG\) (alt. s, BC // AD)
\(\angle AOG = \angle ADC\) (corr. s, CD // BE)
\(\therefore \angle CBE = \angle CDA\)
\(\therefore \Delta BCE \cong \Delta DCA\) (A.A.S.)

Graph

Theoryif the angle at the circumference is a right angle, then the chord that subtends the angle is a diameter

ExamplesIn the figure, chords AC and BD intersect at E. \(\angle BDA = \angle BDC = 45^\circ \).
Find \(\angle ABC\). 
Solutions\(\begin{array}{1}\angle ADC = \angle BDA + \angle BDC\\ = 45^\circ + 45^\circ \\ = 90^\circ \end{array}\)
\(\therefore\) AC is a diameter. (converse of in semicircle)
\(\therefore \angle ABC = \underline{\underline {90^\circ }} \) ( in semicircle)