D-a-d D.a.d Draws a Circle

An axiom is an established or accustomed principle. For this section, the following are accepted every bit axioms.

Theorems (EMBJB)

A theorem is a hypothesis (proposition) that can be shown to be true by accepted mathematical operations and arguments. A proof is the procedure of showing a theorem to exist correct.

The converse of a theorem is the reverse of the hypothesis and the determination. For example, given the theorem "if \(A\), then \(B\)", the converse is "if \(B\), then \(A\)".

Perpendicular line from circumvolve centre bisects chord

If a line is drawn from the centre of a circle perpendicular to a chord, then it bisects the chord.

(Reason: \(\perp\) from eye bisects chord)

Circle with centre \(O\) and line \(OP\) perpendicular to chord \(AB\).

\(AP = Pb\)

Describe \(OA\) and \(OB\).

In \(\triangle OPA\) and in \(\triangle OPB\), \[\begin{array}{rll} OA^2 &= OP^2 + AP^2 & \text{(Pythagoras)} \\ OB^2 &= OP^2 + BP^2 & \text{(Pythagoras)} \stop{array}\] and \[\brainstorm{array}{rll} OA &= OB & \text{(equal radii)} \\ \therefore AP^2 &= BP^2 & \\ \therefore AP &= BP & \end{array}\] Therefore \(OP\) bisects \(AB\).

Alternative proof:

In \(\triangle OPA\) and in \(\triangle OPB\), \[\begin{array}{rll} O\chapeau{P}A &= O\hat{P}B & (\text{given } OP \perp AB) \\ OA &= OB & \text{(equal radii)} \\ OP &= OP & \text{(common side)} \\ \therefore \triangle OPA & \equiv \triangle OPB & \text{(RHS)} \\ \therefore AP &= Pb & \finish{assortment}\] Therefore \(OP\) bisects \(AB\).

(PROOF Non FOR EXAMS) Antipodal: Line from circle eye to mid-point of chord is perpendicular

If a line is drawn from the center of a circumvolve to the mid-point of a chord, then the line is perpendicular to the chord.

(Reason: line from centre to mid-indicate \(\perp\))

Circle with centre \(O\) and line \(OP\) to mid-point \(P\) on chord \(AB\).

\(OP \perp AB\)

Describe \(OA\) and \(OB\).

In \(\triangle OPA\) and in \(\triangle OPB\), \[\brainstorm{assortment}{rll} OA &= OB & \text{(equal radii)} \\ AP &= Pb & \text{(given)} \\ OP &= OP & \text{(common side)} \\ \therefore \triangle OPA & \equiv \triangle OPB & \text{(SSS)} \\ \therefore O\lid{P}A &= O\hat{P}B & \\ \text{and } O\lid{P}A + O\chapeau{P}B &= \text{180}\text{°} & (\bending \text{ on str. line}) \\ \therefore O\hat{P}A = O\chapeau{P}B &= \text{90}\text{°} & \end{assortment}\] Therefore \(OP \perp AB\).

Theorem: Perpendicular bisector of chord passes through circumvolve centre

If the perpendicular bisector of a chord is drawn, then the line will laissez passer through the eye of the circle.

(Reason: \(\perp\) bisector through centre)

Circle with mid-point \(P\) on chord \(AB\).

Line \(QP\) is drawn such that \(Q\hat{P}A = Q\hat{P}B = \text{ninety}\text{°}\).

Line \(RP\) is fatigued such that \(R\chapeau{P}A = R\lid{P}B = \text{90}\text{°}\).

Circle centre \(O\) lies on the line \(PR\)

Describe lines \(QA\) and \(QB\).

Draw lines \(RA\) and \(RB\).

In \(\triangle QPA\) and in \(\triangle QPB\), \[\brainstorm{array}{rll} AP &= Pb & \text{(given)} \\ QP &= QP & \text{(common side)} \\ Q\hat{P}A = Q\lid{P}B &= \text{90}\text{°} & \text{(given)} \\ \therefore \triangle QPA & \equiv \triangle QPB & \text{(SAS)} \\ \therefore QA &= QB & \finish{assortment}\]

Similarly it can be shown that in \(\triangle RPA\) and in \(\triangle RPB\), \(RA = RB\).

We conclude that all the points that are equidistant from \(A\) and \(B\) volition lie on the line \(PR\) extended. Therefore the centre \(O\), which is equidistant to all points on the circumference, must likewise lie on the line \(PR\).

Worked case one: Perpendicular line from circle eye bisects chord

Given \(OQ \perp PR\) and \(PR = viii\) units, determine the value of \(10\).

Use theorems and the given information to observe all equal angles and sides on the diagram

\(PQ = QR = iv \qquad (\perp \text{ from center bisects chord})\)

Solve for \(x\)

In \(\triangle OQP\): \[\brainstorm{array}{rll} PQ &= 4 & (\perp \text{ from middle bisects chord}) \\ OP^2 &= OQ^2 + QP^ii & (\text{Pythagoras}) \\ 5^ii &= x^2 + four^two & \\ \therefore x^ii &= 25 - xvi & \\ x^2 &= ix & \\ x &= iii & \finish{array}\]

Write the final answer

\(x = iii\) units.

Perpendicular line from middle bisects chord

Textbook Exercise 8.1

In the circumvolve with eye \(O\), \(OQ \perp PR\), \(OQ = 4\) units and \(PR=10\). Determine \(x\).

\[\brainstorm{assortment}{rll} PR &= 10 & (\perp \text{ given }) \\ \therefore PQ &= 5 & (\perp \text{ from centre bisects chord}) \\ OP^two &= OQ^ii + QP^2 & (\text{Pythagoras}) \\ x^2 &= 4^2 + 5^2 & \\ \therefore x^ii &= 25 + 16 & \\ ten^2 &= 41 & \\ 10 &= \sqrt{41} & \end{assortment}\]

In the circle with center \(O\) and radius \(= x\) units, \(OQ \perp PR\) and \(PR=viii\). Determine \(10\).

\[\begin{assortment}{rll} PR &= 8 & (\perp \text{ given }) \\ \therefore PQ &= 4 & (\perp \text{ from middle bisects chord}) \\ OP^2 &= OQ^2 + QP^2 & (\text{Pythagoras}) \\ ten^2 &= 10^2 + 4^2 & \\ \therefore ten^ii &= 100 - 16 & \\ x^2 &= 84 & \\ 10 &= \sqrt{84} & \end{assortment}\]

In the circumvolve with centre \(O\), \(OQ \perp PR\), \(PR = 12\) units and \(SQ = two\) units. Determine \(x\).

\[\begin{array}{rll} SO &= x -ii & \\ OP^ii &= OS^ii + SP^2 & (\text{Pythagoras}) \\ ten^2 &= (x-2)^2 + vi^2 & \\ x^2 &= x^2 - 4x + 4 + half-dozen^2 & \\ 4x &= twoscore & \\ \therefore x &= x & \end{array}\]

In the circle with centre \(O\), \(OT \perp SQ\), \(OT \perp PR\), \(OP = 10\) units, \(ST = 5\) units and \(PU = 8\) units. Make up one's mind \(TU\).

\[\begin{assortment}{rll} \text{In } \triangle POU, \quad OP^2 &= OU^2 + PU^2 & (\text{Pythagoras}) \\ ten^2 &= OU^2 + viii^two & \\ 100 - 64 &= OU^2 & \\ \therefore OU^ii &= 36 & \\ \therefore OU &= six & \\ \text{In } \triangle QTO, \quad QO^2 &= OT^two + TQ^two & (\text{Pythagoras}) \\ 10^2 &= OT^2 + v^2 & \\ 100 - 25 &= OT^2 & \\ \therefore OT^2 &= 75 & \\ \therefore OT &= \sqrt{75} & \\ \therefore TU &= OT - OU & \\ &= \sqrt{75} - 6 & \\ &= \text{2,66} & \end{array}\]

In the circumvolve with centre \(O\), \(OT \perp QP\), \(OS \perp PR\), \(OT = five\) units, \(PQ = 24\) units and \(PR = 25\) units. Make up one's mind \(OS = 10\).

\[\begin{assortment}{rll} \text{In } \triangle QTO, \quad QO^2 &= OT^2 + QT^ii & (\text{Pythagoras}) \\ QO^2 &= 5^2 + 12^2 & \\ &= 25 + 144 & \\ \therefore QO^2 &= 169 & \\ \therefore QO &= 13 & \\ \text{In } \triangle OSR, \quad OR^2 &= SR^2 + OS^2 & (\text{Pythagoras}) \\ 13^ii &= \text{12,5}^2 + OS^ii & \\ \therefore OS^2 &= \text{12,75} & \\ \therefore Bone &= \text{3,6} & \end{assortment}\]

Angles subtended by an arc at the centre and the circumference of a circle

1. Measure out angles \(x\) and \(y\) in each of the post-obit graphs:

2. Complete the table:

3. Employ your results to brand a theorize well-nigh the relationship between angles subtended by an arc at the centre of a circumvolve and angles at the circumference of a circumvolve.
4. Now depict three of your own like diagrams and measure out the angles to check your conjecture.

Theorem: Angle at the centre of a circumvolve is twice the size of the angle at the circumference

If an arc subtends an bending at the eye of a circle and at the circumference, then the angle at the middle is twice the size of the angle at the circumference.

(Reason: \(\angle \text{ at centre } = two \angle \text{ at circum.}\))

Circle with eye \(O\), arc \(AB\) subtending \(A\lid{O}B\) at the middle of the circle, and \(A\hat{P}B\) at the circumference.

\(A\hat{O}B= 2A\hat{P}B\)

Draw \(PO\) extended to \(Q\) and let \(A\chapeau{O}Q = \hat{O}_1\) and \(B\hat{O}Q = \hat{O}_2\).

\[\begin{array}{rll} \chapeau{O}_1&= A\hat{P}O + P\hat{A}O & (\text{ext. } \bending \triangle = \text{sum int. opp. } \angle \text{s} ) \\ \text{and } A\hat{P}O &= P\hat{A}O & (\text{equal radii, isosceles } \triangle APO)\\ \therefore \hat{O}_1&= A\lid{P}O + A\lid{P}O & \\ \hat{O}_1&= 2A\hat{P}O & \cease{assortment}\]

Similarly, we can likewise prove that \(\hat{O}_2 = 2B\hat{P}O\).

For the starting time two diagrams shown higher up nosotros accept that: \[\begin{array}{rll} A\hat{O}B &= \chapeau{O}_1 + \chapeau{O}_2 & \\ &= 2A\hat{P}O + 2B\chapeau{P}O & \\ &= ii(A\hat{P}O + B\hat{P}O) & \\ \therefore A\hat{O}B &= two(A\hat{P}B) & \end{array}\] And for the concluding diagram: \[\brainstorm{array}{rll} A\hat{O}B &= \hat{O}_2 - \lid{O}_1 & \\ &= 2B\chapeau{P}O - 2A\lid{P}O & \\ &= 2(B\chapeau{P}O - A\hat{P}O) & \\ \therefore A\chapeau{O}B &= ii(A\lid{P}B) & \end{assortment}\]

Worked instance ii: Angle at the center of circle is twice angle at circumference

Given \(HK\), the diameter of the circumvolve passing through centre \(O\).

Use theorems and the given data to find all equal angles and sides on the diagram

Solve for \(a\)

In \(\triangle HJK\): \[\begin{array}{rll} H\lid{O}K &= \text{180}\text{°} & (\angle \text{ on str. line)} \\ &= 2a & (\angle \text{ at centre } = ii \bending \text{ at circum.}) \\ \therefore 2a &= \text{180}\text{°} & \\ a &= \frac{\text{180}\text{°}}{2} & \\ &= \text{90}\text{°} & \finish{array}\]

Determination

The diameter of a circle subtends a correct angle at the circumference (angles in a semi-circle).

Bending at the centre of circle is twice bending at circumference

Textbook Exercise 8.2

\[\begin{assortment}{rll} b &= two \times \text{45}\text{°} & (\bending \text{ at centre } = 2 \angle \text{ at circum.}) \\ \therefore b &= \text{90}\text{°} & \end{array}\]

\[\begin{array}{rll} c &= \frac{i}{2} \times \text{45}\text{°} & (\angle \text{ at centre } = 2 \angle \text{ at circum.}) \\ \therefore c &= \text{22,5}\text{°} & \stop{array}\]

\[\begin{array}{rll} d &= ii \times \text{100}\text{°} & (\angle \text{ at centre } = ii \angle \text{ at circum.}) \\ \therefore d &= \text{200}\text{°} & \end{array}\]

\[\begin{array}{rll} east &= \text{100}\text{°} - \text{90}\text{°} - \text{35}\text{°} & (\angle \text{ in semi circle} ) \\ \therefore e &= \text{55}\text{°} & \cease{array}\]

\[\begin{array}{rll} f &= \frac{1}{2} \times \text{240}\text{°} & (\angle \text{ at centre } = 2 \angle \text{ at circum.}) \\ \therefore f &= \text{120}\text{°} & \end{array}\]

Subtended angles in the same segment of a circle

1. Measure angles \(a\), \(b\), \(c\), \(d\) and \(east\) in the diagram below:

   

2. Choose whatsoever ii points on the circumference of the circle and label them \(A\) and \(B\).

3. Draw \(AP\) and \(BP\), and measure \(A\hat{P}B\).

4. Draw \(AQ\) and \(BQ\), and measure \(A\hat{Q}B\).

5. What do you observe? Make a conjecture about these types of angles.

Theorem: Subtended angles in the same segment of a circumvolve are equal

If the angles subtended by a chord of the circle are on the same side of the chord, then the angles are equal.

(Reason: \(\angle\)s in same seg.)

Circle with centre \(O\), and points \(P\) and \(Q\) on the circumference of the circumvolve. Arc \(AB\) subtends \(A\lid{P}B\) and \(A\chapeau{Q}B\) in the same segment of the circle.

\(A\hat{P}B = A\chapeau{Q}B\)

\[\brainstorm{assortment}{rll} A\hat{O}B &= two A\hat{P}B & (\angle \text{ at centre } = ii \angle \text{ at circum.}) \\ A\hat{O}B &= ii A\hat{Q}B & (\angle \text{ at centre } = 2 \angle \text{ at circum.}) \\ \therefore 2 A\lid{P}B &= 2 A\hat{Q}B & \\ A\hat{P}B &= A\hat{Q}B & \stop{array}\]

Equal arcs subtend equal angles

From the theorem above we tin can deduce that if angles at the circumference of a circumvolve are subtended by arcs of equal length, and so the angles are equal. In the figure below, notice that if we were to move the two chords with equal length closer to each other, until they overlap, we would have the aforementioned situation as with the theorem in a higher place. This shows that the angles subtended by arcs of equal length are too equal.

(PROOF Not FOR EXAMS) Converse: Concyclic points

If a line segment subtends equal angles at two other points on the same side of the line segment, and then these four points are concyclic (lie on a circumvolve).

Line segment \(AB\) subtending equal angles at points \(P\) and \(Q\) on the same side of the line segment \(AB\).

\(A\), \(B\), \(P\) and \(Q\) lie on a circle.

Proof by contradiction:

Points on the circumference of a circle: we know that there are merely 2 possible options regarding a given point — it either lies on circumference or it does not.

We will assume that signal \(P\) does not lie on the circumference.

We depict a circle that cuts \(AP\) at \(R\) and passes through \(A\), \(B\) and \(Q\). \[\begin{array}{rll} A\hat{Q}B &= A\hat{R}B & (\angle \text{south in same seg.}) \\ \text{only } A\chapeau{Q}B &= A\hat{P}B & (\text{given}) \\ \therefore A\chapeau{R}B &= A\lid{P}B & \\ \text{only } A\lid{R}B &= A\lid{P}B + R\hat{B}P & (\text{ext. } \bending \triangle = \text{sum int. opp.}) \\ \therefore R\hat{B}P &= \text{0}\text{°} & \terminate{array}\] Therefore the assumption that the circumvolve does not pass through \(P\) must be false.

We tin can conclude that \(A\), \(B\), \(Q\) and \(P\) lie on a circle (\(A\), \(B\), \(Q\) and \(P\) are concyclic).

Worked instance three: Concyclic points

Given \(FH \parallel EI\) and \(E\hat{I}F = \text{fifteen}\text{°}\), determine the value of \(b\).

Utilise theorems and the given information to find all equal angles on the diagram

Solve for \(b\)

\[\begin{array}{rll} H\lid{F}I &= \text{15}\text{°} & (\text{alt. } \bending, FH \parallel EI) \\ \text{and } b &= H\chapeau{F}I & (\angle \text{due south in same seg.}) \\ \therefore b &= \text{fifteen}\text{°} & \end{assortment}\]

Subtended angles in the same segment

Textbook Exercise 8.3

\[\begin{assortment}{rll} a &= \text{21}\text{°} & (\angle \text{s in aforementioned seg.}) \end{array}\]

\[\brainstorm{assortment}{rll} c &= \text{24}\text{°} & (\angle \text{south in same seg.}) \\ d &= \text{102}\text{°} - \text{24}\text{°} & (\text{ext. } \angle \triangle = \text{sum int. opp.}) \\ \therefore d &= \text{78}\text{°} & \\ \terminate{array}\]

\[\begin{array}{rll} d &= \hat{North} & (\text{alt. } \angle, PO \parallel QN ) \\ \hat{North} &= \frac{one}{2} \times \text{17}\text{°} & (\angle \text{ at heart } = 2 \bending \text{ at circum.}) \\ \lid{O} &= \hat{North} & (\angle \text{southward in same seg.}) \\ \text{17}\text{°} &= \chapeau{O} + d & ( \text{ ext. bending of } \triangle) \\ \therefore 2d &= \text{17}\text{°} & \\ \therefore d &= \text{8,5}\text{°} & (\text{alt. } \angle, PO \parallel QN ) \end{array}\]

Given \(T\hat{V}S = S\chapeau{V}R\), make up one's mind the value of \(e\).

\[\brainstorm{array}{rll} \text{In } \triangle TRV, \chapeau{T} &= \text{180}\text{°} - (\text{80}\text{°} + \text{xxx}\text{°}) & (\angle \text{s sum of } \triangle) \\ \therefore \hat{T} &= \text{70}\text{°} & \\ \therefore e &= \text{fifteen}\text{°} + \text{70}\text{°} & \\ &= \text{85}\text{°} & \\ \end{array}\]

Is \(TV\) a bore of the circle? Explicate your respond.

No, since \(\text{45}\text{°} + \text{35}\text{°} \ne \text{90}\text{°}\)

Given circle with centre \(O\), \(WT = TY\) and \(X\hat{Westward}T = \text{35}\text{°}\). Determine \(f\).

\[\begin{array}{rll} \text{In } \triangle WTZ &\text{ and in } \triangle YTZ, \\ WT &= YT & (\text{ given }) \\ ZT &= ZT & (\text{ mutual side }) \\ Y\hat{T}Z = W\hat{T}Z &= \text{90}\text{°} & (\text{ line from circle heart to mid-point }) \\ \therefore T\hat{Z}Y &= T\hat{Z}W & (\text{ SAS }) \\ T\hat{Z}Y = T\hat{Z}Due west &= f & \\ \text{And } T\lid{Z}Y &= \text{35}\text{°} & (\bending \text{s in same seg.}) \\ \therefore T\hat{Z}Due west = f &= \text{35}\text{°} & \end{array}\]

Circadian quadrilaterals

Cyclic quadrilaterals are quadrilaterals with all 4 vertices lying on the circumference of a circumvolve (concyclic).

Cyclic quadrilaterals

Consider the diagrams given beneath:

1. Consummate the following:

   \(ABCD\) is a circadian quadrilateral considering \(\ldots \ldots\)

2. Complete the table:

   	Circle \(\text{1}\)	Circumvolve \(\text{2}\)	Circle \(\text{3}\)
   \(\chapeau{A} =\)
   \(\hat{B} =\)
   \(\hat{C} =\)
   \(\hat{D} =\)
   \(\lid{A} + \hat{C} =\)
   \(\hat{B} + \lid{D} =\)

3. Use your results to make a conjecture about the relationship between angles of cyclic quadrilaterals.

Theorem: Contrary angles of a cyclic quadrilateral

The opposite angles of a cyclic quadrilateral are supplementary.

(Reason: opp. \(\angle\)s cyclic quad.)

Circle with centre \(O\) with points \(A, B, P\) and \(Q\) on the circumference such that \(ABPQ\) is a circadian quadrilateral.

\(A\lid{B}P + A\chapeau{Q}P = \text{180}\text{°}\) and \(Q\chapeau{A}B + Q\hat{P}B = \text{180}\text{°}\)

Draw \(AO\) and \(OP\). Label \(\hat{O}_1\) and \(\hat{O}_2\). \[\brainstorm{assortment}{rll} \hat{O}_1 &= 2A\hat{B}P & (\bending\text{ at centre} = 2\angle \text{ at circum.})\\ \hat{O}_2 &= 2A\hat{Q}P & (\bending\text{ at centre} = ii\bending \text{ at circum.})\\ \text{and } \hat{O}_1 + \hat{O}_2 &= \text{360}\text{°} & (\angle\text{southward around a point}) \\ \therefore 2A\hat{B}P + 2A\hat{Q}P &= \text{360}\text{°} & \\ A\hat{B}P + A\hat{Q}P &= \text{180}\text{°} & \terminate{array}\] Similarly, nosotros can show that \(Q\hat{A}B + Q\hat{P}B = \text{180}\text{°}\).

Converse: interior opposite angles of a quadrilateral

If the interior reverse angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.

Exterior angle of a cyclic quadrilateral

If a quadrilateral is circadian, then the outside angle is equal to the interior reverse angle.

Worked example iv: Opposite angles of a circadian quadrilateral

Given the circumvolve with centre \(O\) and cyclic quadrilateral \(PQRS\). \(SQ\) is drawn and \(S\hat{P}Q = \text{34}\text{°}\). Decide the values of \(a\), \(b\) and \(c\).

Use theorems and the given data to find all equal angles on the diagram

Solve for \(b\)

\[\begin{assortment}{rll} S\chapeau{P}Q + c &= \text{180}\text{°} & (\text{opp. } \angle \text{s cyclic quad supp.}) \\ \therefore c &= \text{180}\text{°} - \text{34}\text{°} & \\ &= \text{146}\text{°} & \finish{array}\] \[\begin{array}{rll} a &= \text{ninety}\text{°} & (\angle \text{ in semi circumvolve}) \end{assortment}\] In \(\triangle PSQ\): \[\begin{assortment}{rll} a + b + \text{34}\text{°} &= \text{180}\text{°} & (\angle \text{ sum of } \triangle) \\ \therefore b &= \text{180}\text{°} - \text{90}\text{°} - \text{34}\text{°} & \\ &= \text{56}\text{°} & \terminate{array}\]

Methods for proving a quadrilateral is cyclic

There are three ways to prove that a quadrilateral is a circadian quadrilateral:

	Method of proof	Reason
	If \(\hat{P} + \hat{R} = \text{180}\text{°}\) or \(\chapeau{S} + \lid{Q} = \text{180}\text{°}\), so \(PQRS\) is a cyclic quad.	opp. int. angles suppl.
	If \(\hat{P} = \lid{Q}\) or \(\hat{Due south} = \chapeau{R}\), then \(PQRS\) is a cyclic quad.	angles in the aforementioned seg.
	If \(T\chapeau{Q}R = \hat{S}\), then \(PQRS\) is a cyclic quad.	ext. angle equal to int. opp. bending

Worked case 5: Proving a quadrilateral is a cyclic quadrilateral

Prove that \(ABDE\) is a cyclic quadrilateral.

Use theorems and the given information to find all equal angles on the diagram

Bear witness that \(ABDE\) is a circadian quadrilateral

\[\begin{array}{rll} D\lid{B}C &= \text{90}\text{°} & (\angle \text{ in semi circle}) \\ \text{and } \hat{E} &= \text{90}\text{°} & (\text{given}) \\ \therefore D\hat{B}C &= \hat{E} & \\ \therefore ABDE \text{ is a circadian} & \text{quadrilateral} & \text{(ext.\@ \(\angle\) equals int.\@ opp.\@ \(\angle\))} \end{array}\]

Cyclic quadrilaterals

Textbook Exercise 8.4

\[\begin{array}{rll} a + \text{87}\text{°} &= \text{180}\text{°} & (\text{opp. angles of cyclic quad. supp. }) \\ \therefore a &= \text{93}\text{°} & \\ b + \text{106}\text{°} &= \text{180}\text{°} & (\text{opp. angles of cyclic quad. supp. }) \\ \therefore b &= \text{74}\text{°} & \end{array}\]

\[\begin{array}{rll} a &= H\chapeau{I}J & \angle (\text{ ext. angle cyclic quad = int. opp }) \\ &= \text{114}\text{°} & \end{array}\]

\[\brainstorm{array}{rll} \hat{W} + \text{86}\text{°} &= \text{180}\text{°} & (\text{opp. angles of cyclic quad. supp. }) \\ \therefore \lid{W} &= \text{94}\text{°} & \\ a + \lid{Due west} + \text{57}\text{°} &= \text{180}\text{°} & (\text{angles sum of } \triangle) \\ \therefore a &= \text{29}\text{°} & \end{array}\]

\[\brainstorm{assortment}{rll} A\hat{Chiliad}B &= \text{32}\text{°} + D\lid{B}C & (\text{ext bending of } \triangle = \text{sum int. opp. angles}) \\ \therefore \text{72}\text{°} &= \text{32}\text{°} + D\chapeau{B}C & \\ \therefore D\lid{B}C &= \text{forty}\text{°} & (\text{angles sum of } \triangle) \\ \therefore D\hat{B}C &= D\lid{A}C & \\ \text{Therefore } ABCD &= \text{ is cyclic quad. } & ( \text{ angles in same seg.}) \end{assortment}\]

\[\brainstorm{array}{rll} \text{In } \triangle ABD, \quad A\hat{B}D = A\hat{D}B &= \text{35}\text{°} & (\angle \text{s opp. equal sides }) \\ \therefore \text{35}\text{°} + \text{35}\text{°} + D\hat{A}B &= \text{180}\text{°} & (\angle \text{south sum of } \triangle) \\ \therefore D\hat{A}B &= \text{110}\text{°} & (\text{angles sum of } \triangle) \\ \text{And } D\hat{A}B + D\hat{C}B &= \text{180}\text{°} & \\ \text{Therefore } ABCD &= \text{ is cyclic quad. } & ( \text{ opp. int. angles supp.}) \finish{array}\]

Tangent line to a circle

A tangent is a line that touches the circumference of a circle at only one place. The radius of a circle is perpendicular to the tangent at the point of contact.

Theorem: Two tangents drawn from the same point outside a circle

If two tangents are fatigued from the aforementioned betoken outside a circle, then they are equal in length.

(Reason: tangents from same betoken equal)

Circle with middle \(O\) and tangents \(PA\) and \(PB\), where \(A\) and \(B\) are the respective points of contact for the ii lines.

\(AP = BP\)

In \(\triangle AOP\) and \(\triangle BOP\), \[\begin{array}{rll} O\hat{A}P = O\hat{B}P &= \text{ninety}\text{°} & (\text{tangent} \perp \text{radius})\\ AO &= BO & (\text{equal radii}) \\ OP &= OP & (\text{common side}) \\ \therefore \triangle AOP &\equiv \triangle BOP & (\text{RHS}) \\ \therefore AP &= BP & \end{array}\]

Worked case 6: Tangents from the same point outside a circle

In the diagram below \(AE = \text{5}\text{ cm}\), \(Air conditioning = \text{8}\text{ cm}\) and \(CE = \text{ix}\text{ cm}\). Decide the values of \(a\), \(b\) and \(c\).

Use theorems and the given information to observe all equal angles on the diagram

Solve for \(a\), \(b\) and \(c\)

\[\begin{assortment}{rll} AB = AF&= a & (\text{tangents from } A) \\ EF = ED&= c & (\text{tangents from } E) \\ CB = CD&= b & (\text{tangents from } C) \\ \therefore AE = a + c &= 5 & \\ \text{and } Ac = a + b &= eight & \\ \text{and } CE = b + c &= 9 & \stop{assortment}\]

Solve for the unknown variables using simultaneous equations

\[\begin{assortment}{rll} a + c &= v & \ldots (1)\\ a + b &= eight & \ldots (two)\\ b + c &= ix & \ldots (iii) \end{assortment}\]

Subtract equation \((1)\) from equation \((two)\) then substitute into equation \((three)\):

\[\begin{array}{rll} (2) - (1) \quad b-c &= 8-5 & \\ &= three & \\ \therefore b &= c + 3 & \\ \text{Substitute into } (3) \quad c + iii + c &= ix & \\ 2c &= 6 & \\ c &= iii & \\ \therefore a &= 2 & \\ \text{ and } b &= 6 & \end{array}\]

Tangents to a circle

Textbook Do 8.v

\[\begin{assortment}{rll} Hi &= HG & (\text{tangents same pt. }) \\ \text{In } \triangle HIJ, \quad d^two &= eight^2 + five^ii & (\text{Pythagoras, radius perp. tangent }) \\ \therefore d^2 &= 89 & \\ \therefore d &= \text{9,4}\text{ cm} & \end{array}\]

\[\begin{assortment}{rll} LM = LK &= 6 & (\text{tangents same pt. }) \\ LN &= \text{7,v}\text{ cm} & (\text{ given }) \\ \therefore MN = \text{7,5} - \text{6} &= \text{1,v}\text{ cm} & \\ OM &= \text{2}\text{ cm} & (\text{ radius }) \\ \text{In } \triangle OMN, \quad east^two &= ii^2 + (\text{i,5})^ii & (\text{Pythagoras, radius perp. tangent }) \\ \therefore e^2 &= \text{6,25} & \\ \therefore e &= \text{2,5}\text{ cm} & \end{array}\]

\(f=\text{3}\text{ cm}\)

Tangent-chord theorem

Consider the diagrams given below:

1. Mensurate the post-obit angles with a protractor and consummate the table:

   	Diagram \(\text{ane}\)	Diagram \(\text{2}\)	Diagram \(\text{iii}\)
   \(A\hat{B}C =\)
   \(\hat{D} =\)
   \(\hat{East} =\)

2. Utilize your results to complete the following: the angle between a tangent to a circle and a chord is \(\ldots \ldots\) to the angle in the alternating segment.

Theorem: Tangent-chord theorem

The angle between a tangent to a circumvolve and a chord fatigued at the point of contact, is equal to the angle which the chord subtends in the alternate segment.

(Reason: tan. chord theorem)

Circumvolve with heart \(O\) and tangent \(SR\) touching the circumvolve at \(B\). Chord \(AB\) subtends \(\lid{P}_1\) and \(\hat{Q}_1\).

1. \(A\hat{B}R = A\hat{P}B\)
2. \(A\hat{B}S = A\hat{Q}B\)

Draw diameter \(BT\) and join \(T\) to \(A\).

Permit \(A\hat{T}B = T_1\). \[\begin{array}{rll} A\lid{B}South + A\chapeau{B}T &= \text{ninety}\text{°} & (\text{tangent} \perp \text{radius}) \\ B\chapeau{A}T &= \text{90}\text{°} & (\angle \text{ in semi circle}) \\ \therefore A\lid{B}T + T_1 &= \text{90}\text{°} & (\bending \text{ sum of } \triangle BAT) \\ \therefore A\hat{B}S &= T_1 & \\ \text{but } Q_1 &= T_1 & (\bending \text{due south in same segment}) \\ \therefore Q_1 &= A\chapeau{B}S & \cease{array}\] \[\begin{assortment}{rll} A\hat{B}S + A\hat{B}R &= \text{180}\text{°} & (\angle \text{s on str. line}) \\ \hat{Q}_1 + \hat{P}_1 &= \text{180}\text{°} & (\text{opp. } \bending \text{s circadian quad. supp.}) \\ \therefore A\hat{B}S + A\hat{B}R &= Q_1 + P_1 & \\ \text{and } A\hat{B}S &= Q_1 & \\ \therefore A\hat{B}R &= P_1 & \end{array}\]

Worked example 7: Tangent-chord theorem

Determine the values of \(h\) and \(s\).

Employ theorems and the given information to find all equal angles on the diagram

Solve for \(h\)

\[\begin{array}{rll} O\hat{Q}S &= S\hat{R}Q & (\text{tangent chord theorem}) \\ h + \text{xx}\text{°} &= 4h - \text{seventy}\text{°} & \\ \text{90}\text{°} &= 3h & \\ \therefore h &= \text{thirty}\text{°} & \end{array}\]

Solve for \(s\)

\[\begin{assortment}{rll} P\hat{Q}R &= Q\lid{Southward}R & (\text{tangent chord theorem}) \\ s &= 4h & \\ &= iv(\text{30}\text{°}) & \\ &= \text{120}\text{°} & \end{array}\]

Tangent-chord theorem

Textbook Exercise 8.half dozen

\[\begin{array}{rll} a &= \text{33}\text{°} & (\text{tangent-chord }) \\ b &= \text{33}\text{°} & (\text{alt. angles, } OP \parallel SR ) \end{array}\]

\[\begin{array}{rll} c &= \text{72}\text{°} & (\text{tangent-chord}) \\ d &= \dfrac{\text{180}\text{°} - \text{72}\text{°} }{2} & (\text{isosceles triangle}) \\ &= \text{54}\text{°} & (\text{alt. angles, } OP \parallel SR ) \stop{array}\]

\[\begin{array}{rll} f &= \text{38}\text{°} & (\text{tangent-chord}) \\ g &= \text{47}\text{°} & (\text{tangent-chord}) \end{array}\]

\[\begin{array}{rll} \chapeau{O}_1 = \chapeau{Q}_1 &= \text{66}\text{°} & (\text{isosceles, tangent-chord}) \\ \therefore l &= \text{180}\text{°} - 2 \times \text{66}\text{°} & (\text{angles sum } \triangle) \\ &= \text{48}\text{°} & \end{array}\]

\[\brainstorm{array}{rll} i &= \text{180}\text{°} - \text{101}\text{°} - \text{39}\text{°} & (\angle \text{south on str. line }) \\ \therefore i &= \text{40}\text{°} & \\ j &= \text{101}\text{°} & (\text{ tangent-chord}) \\ k = i &= \text{40}\text{°} & (\text{ tangent-chord}) \end{array}\]

\[\brainstorm{assortment}{rll} n &= \text{34}\text{°} & (\text{ tangent-chord}) \\ o &= \text{180}\text{°} - \text{90}\text{°} - \text{34}\text{°} & (\text{angles sum } \triangle) \\ \therefore o &= \text{56}\text{°} & \\ m &= \text{56}\text{°} & (\text{ tangent-chord}) \end{array}\]

\[\brainstorm{array}{rll} q &= \text{52}\text{°} & (\text{ tangent-chord}) \\ p &= \text{xc}\text{°} - \text{52}\text{°} & (\text{tangent perp. radius }) \\ \therefore p &= \text{38}\text{°} & \\ r &= \text{90}\text{°} & ( \angle \text{ in semi-circumvolve}) \terminate{array}\]

\(O\) is the centre of the circle and \(SPT\) is a tangent, with \(OP \perp ST\). Determine \(a\), \(b\) and \(c\), giving reasons.

\[\begin{assortment}{rll} a &= \text{xc}\text{°} - \text{64}\text{°} & (\text{tangent perp. radius}) \\ &= \text{26}\text{°} & \\ b &= \text{64}\text{°} & (\text{tangent chord}) \\ c &= 2 \times \text{64}\text{°} & (\angle\text{ at centre} = ii\angle \text{ at circum.}) \\ &= \text{128}\text{°} & \\ \end{assortment}\]

\(PAL\) is a tangent to the circumvolve \(ABC\).

\[\begin{assortment}{rll} \chapeau{A}_1 &= A\chapeau{C}B & (\text{alt. angles}, AP \parallel BC) \\ A\lid{C}B &= A\hat{B}C & (\angle \text{southward opp. equal sides }, AB = AC) \\ \therefore \hat{A}_1 &= A\chapeau{B}C & \\ \text{Therefore } PAL & \text{is a tangent to circle } ABC & (\angle\text{ between line chord} = \angle\text{ in alt. seg.}) \end{array}\]

\(AB\) is a tangent to the circle \(ADP\).

\[\brainstorm{array}{rll} \hat{A}_2 &= \hat{B}_2 & (\text{ given }) \\ \text{And } A\hat{P}B &= \hat{B}_2 & (\text{alt. angles}, AP \parallel BC) \\ \text{Therefore } A\hat{P}B &= \hat{A}_2 ABC & \\ \text{Therefore } AB & \text{is a tangent to circle } ADP & (\angle\text{ between line chord} = \angle\text{ in alt. seg.}) \stop{assortment}\]

Converse: tangent-chord theorem

If a line fatigued through the end point of a chord forms an angle equal to the angle subtended past the chord in the alternate segment, then the line is a tangent to the circumvolve.

(Reason: \(\angle\) between line and chord \(= \angle\) in alt. seg. )

Worked example 8: Applying the theorems

\(BD\) is a tangent to the circle with centre \(O\), with \(BO \perp AD\).

Prove that:

1. \(CFOE\) is a circadian quadrilateral

2. \(FB=BC\)

3. \(\angle A\hat{O}C = 2 B\lid{F}C\)

4. Will \(DC\) exist a tangent to the circle passing through \(C,F,O\) and \(E\)? Motivate your answer.

Show \(CFOE\) is a cyclic quadrilateral by showing contrary angles are supplementary

\[\brainstorm{assortment}{rll} BO & \perp OD & (\text{given}) \\ \therefore F\hat{O}Due east &= \text{ninety}\text{°} & \\ F\hat{C}E &= \text{90}\text{°} & (\angle \text{ in semi circle}) \\ \therefore CFOE & \text{is a cyclic quad.} & (\text{opp. } \angle \text{s suppl.}) \end{array}\]

Prove \(BFC\) is an isosceles triangle

To show that \(FB = BC\) we first prove \(\triangle BFC\) is an isosceles triangle by showing that \(B\hat{F}C = B\hat{C}F\).

\[\begin{array}{rll} B\chapeau{C}F &= C\lid{E}O & (\text{tangent-chord}) \\ C\chapeau{E}O &= B\hat{F}C & (\text{ext. } \angle \text{ cyclic quad. } CFOE) \\ \therefore B\hat{F}C &= B\hat{C}F \\ \therefore FB &= BC & (\triangle BFC \text{ isosceles}) \end{array}\]

Prove \(A\hat{O}C = ii B\hat{F}C\)

\[\brainstorm{array}{rll} A\hat{O}C &= ii A\chapeau{E}C & (\angle \text{ at heart } = 2 \angle \text{ at circum.}) \\ \text{and } A\lid{Eastward}C &= B\lid{F}C & (\text{ext. } \angle \text{ cyclic quad. } CFOE) \\ \therefore A\hat{O}C &= 2 B\hat{F}C \stop{array}\]

Decide if \(DC\) is a tangent to the circle through \(C\), \(F\), \(O\) and \(E\)

Proof past contradiction.

Let us assume that \(DC\) is a tangent to the circle passing through the points \(C\), \(F\), \(O\) and \(East\): \[\begin{array}{rll} \therefore D\hat{C}E = C\chapeau{O}Due east \quad (\text{tangent-chord}) \end{array}\] And using the circle with centre \(O\) and tangent \(BD\) we have that: \[\begin{array}{rll} D\hat{C}E &= C\hat{A}East & (\text{tangent-chord}) \\ \text{simply } C\hat{A}E &= \frac{i}{two} C\lid{O}E & (\angle \text{ at centre} = ii \angle \text{ at circum.}) \\ \therefore D\hat{C}E &\ne C\hat{O}E & \cease{array}\] Therefore our assumption is non correct and nosotros tin can conclude that \(DC\) is not a tangent to the circumvolve passing through the points \(C\), \(F\), \(O\) and \(E\).

Worked instance 9: Applying the theorems

\(FD\) is fatigued parallel to the tangent \(CB\)

Prove that:

1. \(FADE\) is a cyclic quadrilateral

2. \(F\hat{Due east}A = \hat{B}\)

Prove \(FADE\) is a cyclic quadrilateral using angles in the same segment

\[\begin{array}{rll} F\lid{D}C &= D\hat{C}B & (\text{alt. } \bending\text{s } FD \parallel CB) \\ \text{and } D\hat{C}B &= C\hat{A}East & (\text{tangent-chord}) \\ \therefore F\hat{D}C &= C\hat{A}E \\ \therefore FADE & \text{is a circadian quad.} & (\angle \text{s in same seg.}) \end{assortment}\]

Prove \(F\hat{E}A = \hat{B}\)

\[\begin{array}{rll} F\hat{D}A &= \hat{B} & (\text{corresp. } \angle\text{s } FD \parallel CB) \\ \text{and } F\chapeau{E}A &= F\lid{D}A & (\bending \text{s same seg.\@ cyclic quad.\@ } FADE) \\ \therefore F\hat{E}A &= \hat{B} \end{array}\]

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Source: https://intl.siyavula.com/read/maths/grade-11/euclidean-geometry/08-euclidean-geometry-02

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