### Basic properties of circles - (10.2)

#### Understand the angle properties of a circle

• 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

• In 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.

• $$\begin{array}{1} \angle BOC = 2 \times 42^\circ \\ = {84^\circ {}}\end{array}$$ ( $$\angle$$at centre =2$$\angle$$at ⊙ce)

In $$\Delta$$ OBC,
$$OC = OB$$ (radii)
$$\angle OCB = \angle OBC$$ (base $$\angle$$s, isos.$$\Delta$$)
$$\begin{array}{1} \angle OBC + \angle OCB + \angle BOC = 180^\circ \\2\angle OBC + 84^\circ = 180{^\circ {}}\\2\angle OBC = 96{^\circ {}}\\\angle OBC = {\underline{\underline {48^\circ }} {}}\end{array}$$ ($$\angle$$sum of$$\Delta$$)

In $$\Delta$$ ABC,
$$AC = AB$$ (given)
$$\angle ACB = \angle ABC$$ (base $$\angle$$s, isos.$$\Delta$$)
$$\begin{array}{1} \angle ABC + \angle ACB + \angle BAC = 180^\circ \\2\angle ABC + 42^\circ = 180{^\circ {}}\\2\angle ABC = 138{^\circ {}}\\\angle ABC = 69{^\circ {}}\end{array}$$ ($$\angle$$sum of$$\Delta$$)
$$\begin{array}{1} \angle ABO = 69^\circ - 48^\circ \\ = {\underline{\underline {21^\circ }} ^{}}\end{array}$$

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

• In the figure, chords AC and BD intersect at E. Find $$w + x + y + z$$.
• $$\begin{array}{1}\angle ACD = \angle ABD\\ = {x^{}}\end{array}$$ (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 )
$$\therefore w + x + y + z = 180^\circ$$

• the arcs are proportional to their corresponding angles at the circumference

• In 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 $$\Delta ADC \cong \Delta BCD$$.
• In 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.)

• the angle in a semi-circle is a right angle
• In 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 $$\Delta BCE \cong \Delta DCA$$.
• In BCE and DCA,
$$BE = DA$$ (diameters)
$$\angle BCE = \angle DCA = 90^\circ$$ ( in semi-circle)
$$\angle CBG = \angle AOG$$ (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.)

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

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