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Revision

Chapter 8: Euclidean geometry

  • Sketches are valuable and important tools. Encourage learners to draw accurate diagrams to solve problems.
  • It is important to stress to learners that proportion gives no indication of actual length. It only indicates the ratio between lengths.
  • To prove triangles are similar, we need to show that two angles (AAA) are equal OR three sides in proportion (SSS).
  • Theorems are examinable and are often asked in examinations. It is also important that learners remember the correct construction required for each proof.
  • Notation - emphasize to learners the importance of the correct ordering of letters, as this indicates which angles are equal and which sides are in the same proportion.
  • If a length has to be calculated from a proportion, it helps to re-write the proportion with the unknown length in the top left position.

8.1 Revision (EMCHY)

Types of triangles (EMCHZ)

Name

Diagram

Properties

Scalene

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All sides and angles are different.

Isosceles

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Two sides are equal in length. The angles opposite the equal sides are also equal.

Equilateral

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All three sides are equal in length and all three angles are equal.

Acute-angled

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Each of the three interior angles is less than \(\text{90}\)°.

Obtuse-angled

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One interior angle is greater than \(\text{90}\)°.

Right-angled

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One interior angle is \(\text{90}\)°.

Congruent triangles (EMCJ2)

Condition

Diagram

SSS

(side, side, side)

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\(\triangle ABC \equiv \triangle EDF\)

SAS

(side, incl. angle, side)

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\(\triangle GHI \equiv \triangle JKL\)

AAS

(angle, angle, side)

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\(\triangle MNO \equiv \triangle PQR\)

RHS

(\(\text{90}\)°, hypotenuse, side)

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\(\triangle STU \equiv \triangle VWX\)

Similar triangles (EMCJ3)

Condition

Diagram

AAA

(angle, angle, angle)

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\(\hat{A} = \hat{D}, \enspace \hat{B} = \hat{E}, \enspace \hat{C} = \hat{F}\)

\(\therefore \triangle ABC \enspace ||| \enspace \triangle DEF\)

SSS

(sides in prop.)

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\(\frac{MN}{RS} = \frac{ML}{RT} = \frac{NL}{ST}\)

\(\therefore \triangle MNL \enspace ||| \enspace \triangle RST\)

Circle geometry (EMCJ4)

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  • If \(O\) is the centre and \(OM \perp AB\), then \(AM = MB\).
  • If \(O\) is the centre and \(AM = MB\), then \(A\hat{M}O = B\hat{M}O = \text{90}°\).
  • If \(AM = MB\) and \(OM \perp AB\), then \(\Rightarrow MO\) passes through centre \(O\).
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If an arc subtends an angle at the centre of a circle and at the circumference, then the angle at the centre is twice the size of the angle at the circumference.

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Angles at the circumference subtended by arcs of equal length (or by the same arc) are equal.

Cyclic quadrilaterals (EMCJ5)

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If the four sides of a quadrilateral \(ABCD\) are the chords of a circle with centre \(O\), then:

  • \(\hat{A} + \hat{C} = \text{180}°\)

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

  • \(\hat{B} + \hat{D} = \text{180}°\)

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

  • \(E\hat{B}C = \hat{D}\)

    Reason: (ext. \(\angle\) cyclic quad. = int. opp \(\angle\))

  • \(\hat{A}_1 = \hat{A}_2 = \hat{C}\)

    Reason: (vert. opp. \(\angle\)s, ext. \(\angle\) cyclic quad.)

Proving a quadrilateral is cyclic:

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If \(\hat{A} + \hat{C} = \text{180}°\) or \(\hat{B} + \hat{D} = \text{90}´\), then \(ABCD\) is a cyclic quadrilateral.

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If \(\hat{A}_1 = \hat{C}\) or \(\hat{D}_1 = \hat{B}\), then \(ABCD\) is a cyclic quadrilateral.

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If \(\hat{A} = \hat{B}\) or \(\hat{C} = \hat{D}\), then \(ABCD\) is a cyclic quadrilateral.

Tangents to a circle (EMCJ6)

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A tangent is perpendicular to the radius (\(OT \perp ST\)), drawn to the point of contact with the circle.

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If \(AT\) and \(BT\) are tangents to a circle with centre \(O\), then:

  • \(OA \perp AT\) (tangent \(\perp\) radius)
  • \(OB \perp BT\) (tangent \(\perp\) radius)
  • \(TA = TB\) (tangents from same point are equal)
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  • If \(DC\) is a tangent, then \(D\hat{T}A = T\hat{B}A\) and \(C\hat{T}B = T\hat{A}B\).
  • If \(D\hat{T}A = T\hat{B}A\) or \(C\hat{T}B = T\hat{A}B\), then \(DC\) is a tangent touching at \(T\).

The mid-point theorem (EMCJ7)

The line joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the length of the third side.

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Given: \(AD = DB\) and \(AE = EC\), we can conclude that \(DE \parallel BC\) and \(DE = \frac{1}{2}BC\).

Revision

Exercise 8.1

\(MO \parallel NP\) in a circle with centre \(O\). \(M\hat{O}N = \text{60}°\) and \(O\hat{M}P = z\). Calculate the value of \(z\), giving reasons.

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\[\begin{array}{rll} \hat{P} &= \frac{1}{2} M\hat{O}N & (\angle \text{ at centre = twice } \angle \text{ at circumference})\\ &= \text{30}° & \\ \therefore z &= \text{30}° &(\text{alt. } \angle\text{s, } MO \parallel NP) \end{array}\]

\(O\) is the centre of the circle with \(OC = \text{5}\text{ cm}\) and chord \(BC = \text{8}\text{ cm}\).

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Determine the lengths of:

\(OD\)
\[\begin{array}{rll} \text{In } \triangle ODC, \quad OC^2 &= OD^2 + DC^2 &(\text{Pythagoras})\\ 5^2 &= OD^2 + 4^2 & \\ \therefore OD &= \text{3}\text{ cm} & \end{array}\]
\(AD\)
\[\begin{array}{rll} AO &= \text{5}\text{ cm} &\text{(radius)} \\ AD &= AO + OD & \\ &= \text{5} + \text{3} & \\ \therefore AD &= \text{8}\text{ cm} & \\ \end{array}\]
\(AB\)
\[\begin{array}{rll} \text{In } \triangle ABD, \quad AB^2 &= BD^2 + AD^2 &(\text{Pythagoras}) \\ AB^2 &= 4^2 + 8^2 & \\ AB &= \sqrt{80} & \\ \therefore AB &= 4\sqrt{5}\text{cm} & \end{array}\]

\(PQ\) is a diameter of the circle with centre \(O\). \(SQ\) bisects \(P\hat{Q}R\) and \(P\hat{Q}S = a\).

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Write down two other angles that are also equal to \(a\).

\[\begin{array}{rll} R\hat{Q}S &= a & (\text{given } SQ \text{ bisects } P\hat{Q}R) \\ OQ &= OS & (\text{equal radii}) \\ \therefore O\hat{Q}S &= O\hat{S}Q = a & (\text{isosceles } \triangle OQS) \end{array}\]

Calculate \(P\hat{O}S\) in terms of \(a\), giving reasons.

\[\begin{array}{rll} P\hat{O}S &= \text{2}\times P\hat{Q}S&(\angle \text{s at centre and circumference on same chord}) \\ &= \text{2}a & \end{array}\]

Prove that \(OS\) is a perpendicular bisector of \(PR\).

\[\begin{array}{rll} R\hat{Q}S &= Q\hat{S}O = a & (\text{proven}) \\ \therefore QR &\parallel OS & (\text{alt. } \angle \text{s equal}) \\ \therefore \hat{R} &= R\hat{T}S & (\text{alt. } \angle \text{s, } QR \parallel OS)\\ &= \text{90}° &(\hat{R} = \angle \text{ in semi-circle})\\ \therefore PT &= TR & (\perp \text{from centre bisects chord})\\ \therefore OS &\text{ perp. bisector of } PR & \end{array}\]

\(BD\) is a diameter of the circle with centre \(O\). \(AB = AD\) and \(O\hat{C}D = \text{35}°\).

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Calculate the value of the following angles, giving reasons:

\(O\hat{D}C\)
\[\begin{array}{rll} OC &= OD &(\text{equal radii }) \\ \therefore O\hat{D}C &= \text{35}°&(\text{isosceles } \triangle OCD) \end{array}\]
\(C\hat{O}D\)
\[\begin{array}{rll} C\hat{O}D &= \text{180}° - \left( \text{35}° + \text{35}° \right)&(\text{sum } \angle\text{s } \triangle = \text{180}°) \\ &=\text{110}° & \end{array}\]
\(C\hat{B}D\)
\[\begin{array}{rll} C\hat{B}D &= \frac{1}{2} C\hat{O}D&(\angle\text{ at centre } = 2\angle\text{ at circum. } )\\ &= \text{55}° & \end{array}\]
\(B\hat{A}D\)
\[\begin{array}{rll} B\hat{A}D &= \text{90}° & (\angle\text{ in semi-circle}) \end{array}\]
\(A\hat{D}B\)
\[\begin{array}{rll} A\hat{D}B &= A\hat{B}D &(\text{isosceles } \triangle ABD) \\ \therefore A\hat{D}B &= \frac{\text{180}°-\text{90}°}{2}&(\text{sum } \angle\text{s in } \triangle = \text{180}°)\\ &= \text{45}° & \end{array}\]

\(O\) is the centre of the circle with diameter \(AB\). \(CD \perp AB\) at \(P\) and chord \(DE\) intersects \(AB\) at \(F\).

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Prove the following:

\(C\hat{B}P = D\hat{B}P\)
\[\begin{array}{rll} \text{In } \triangle CBP &\text{ and } \triangle DBP\text{:}& \\ CP &= DP& (OP \perp CD)\\ C\hat{P}B &= D\hat{P}B = \text{90}° & (\text{given}) \\ BP &= BP & (\text{common})\\ \therefore \triangle CBP &\equiv \triangle DBP &(\text{SAS} ) \\ \therefore C\hat{B}P &= D\hat{B}P & (\triangle CBP \equiv \triangle DBP) \end{array}\\\]
\(C\hat{E}D = 2 C\hat{B}A\)
\[\begin{array}{rll} C\hat{E}D &= C\hat{B}D& (\angle\text{s on chord } CD)\\ \text{But } C\hat{B}A &= D\hat{B}A & (\triangle CBP \equiv \triangle DBP)\\ \therefore C\hat{E}D &= 2 C\hat{B}A \end{array}\]
\(A\hat{B}D = \frac{1}{2} C\hat{O}A\)
\[\begin{array}{rll} D\hat{B}A &= C\hat{B}A & (\triangle CBP \equiv \triangle DBP)\\ C\hat{B}A &= \frac{1}{2} C\hat{O}A &(\angle\text{ at centre } = 2\angle\text{ at circum. } )\\ \therefore A\hat{B}D &= \frac{1}{2} C\hat{O}A & \end{array}\]

\(QP\) in the circle with centre \(O\) is extended to \(T\) so that \(PR = PT\). Express \(m\) in terms of \(n\).

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\[\begin{array}{rll} \hat{T} &= m &(PT = PR)\\ \therefore Q\hat{P}R &= \text{2}m &(\text{ext. } \angle \triangle = \text{ sum int. } \angle\text{s})\\ \therefore n &= \text{2}(\text{2}m)& (\angle\text{s at centre and circumference on } QR)\\ n &= 4m & \\ \therefore m &= \frac{1}{4}n & \end{array}\]

In the circle with centre \(O\), \(OR \perp QP\), \(QP = \text{30}\text{ mm}\) and \(RS = \text{9}\text{ mm}\). Determine the length of \(y\).

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\[\begin{array}{rll} \text{In } &\triangle QOS, & \\ QP &= 30 & ( \text{given}) \\ QS &= \frac{1}{2}QP & (\perp \text{ from centre bisects chord}) \\ \therefore QS &= 15 & \\ QO^2 &= OS^2 + QS^2 & ( \text{Pythagoras}) \\ y^2 &= (y-9)^2 + 15^2 & \\ y^2 &= y^2 - 18y + 81 + 225 & \\ \therefore 18y &= 306 & \\ \therefore y &= \text{17}\text{ mm} & \end{array}\]

\(PQ\) is a diameter of the circle with centre \(O\). \(QP\) is extended to \(A\) and \(AC\) is a tangent to the circle. \(BA \perp AQ\) and \(BCQ\) is a straight line.

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Prove the following:

\(P\hat{C}Q = B\hat{A}P\)
\[\begin{array}{rll} P\hat{C}Q &= \text{90}° & (\angle \text{ in semi-circle} ) \\ B\hat{A}Q &= \text{90}° & (\text{given } BA \perp AQ) \\ \therefore P\hat{C}Q &= B\hat{A}Q & \end{array}\]

\(BAPC\) is a cyclic quadrilateral

\[\begin{array}{rll} P\hat{C}Q &= B\hat{A}Q & ( \text{proven} ) \\ \therefore &BAPC \text{ is a cyclic quad. } & ( \text{ext. angle = int. opp. } \angle ) \end{array}\]
\(AB = AC\)
\[\begin{array}{rll} C\hat{P}Q &= A\hat{B}C & ( \text{ext. } \angle \text{ of cyclic quad.}) \\ B\hat{C}P &= C\hat{P}Q + C\hat{Q}P & ( \text{ext. } \angle \text{ of } \triangle) \\ A\hat{C}P &= C\hat{Q}P & ( \text{tangent-chord} ) \\ \therefore B\hat{C}A &= C\hat{P}Q & \\ &= A\hat{B}C & \\ \therefore AB &= AC & (\angle \text{s opp. equal sides}) \end{array}\]

\(TA\) and \(TB\) are tangents to the circle with centre \(O\). \(C\) is a point on the circumference and \(A\hat{T}B = x\).

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Express the following in terms of \(x\), giving reasons:

\(A\hat{B}T\)
\[\begin{array}{rll} A\hat{B}T &= B\hat{A}T & (TA = TB) \\ &= \frac{\text{180}° - x}{2} & ( \text{sum } \angle \text{s of } \triangle TAB) \\ &= \text{90}° - \frac{x}{2} \end{array}\]
\(O\hat{B}A\)
\[\begin{array}{rll} O\hat{B}T &= \text{90}° & (\text{tangent } \perp \text{ radius}) \\ \therefore O\hat{B}A &= \text{90}° - \left( \text{90}° - \frac{x}{2} \right) & \\ &= \frac{x}{2} \end{array}\]
\(\hat{C}\)
\[\begin{array}{rll} \hat{C} &= A\hat{B}T & (\text{tangent chord}) \\ &= \text{90}° - \frac{x}{2} & \end{array}\]