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8.1 Revision

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End of chapter exercises
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8.2 Ratio and proportion

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)

temp text

Types of triangles (EMCHZ)

Name

Diagram

Properties

Scalene

 009059e85d03a14b0535ccbe43991bb0.png

All sides and angles are different.

Isosceles

 4e56e4dc5efd0035fdbf310a383e261f.png

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

Equilateral

 cf875f8ae40ec9cb7d532401b0e9383a.png

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

Acute-angled

 cc9650766b7bf6798ac6a86ea8cd3cb4.png

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

Obtuse-angled

 501a63891137afb1e0ba73111f218a87.png

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

Right-angled

 d3102e4ac20d4af698e91bd839b26039.png

One interior angle is \(\text{90}\)°.

Congruent triangles (EMCJ2)

Condition

Diagram

SSS

(side, side, side)

f175ae28c9bd1cb3bc8485ef9e02b3a3.png

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

SAS

(side, incl. angle, side)

fc00adae9adde821c7ef8f0586c22565.png

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

AAS

(angle, angle, side)

e3bba55b97ce1cdd09da7470c4a9b372.png

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

RHS

(\(\text{90}\)°, hypotenuse, side)

a68c788e132a1ef6a85e83640bdf9a22.png

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

Similar triangles (EMCJ3)

Condition

Diagram

AAA

(angle, angle, angle)

 519c33b81f7a1f5cf9dce6cfdbde8e21.png

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

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

SSS

(sides in prop.)

 e5764b1ece334ed76648b2f253f8fd5b.png

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

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

Circle geometry (EMCJ4)

a92901b6ecbbf8bb2df1ed58e5c97e2b.png
  • 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 \(O\).
ebabdb37e2b532b93c38ceb17c17ee21.png

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.
6096141afe1ec2ae1ebcfed1d8e4825c.png

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

Cyclic quadrilaterals (EMCJ5)

613598fad88641189a8f83d398281a16.png

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:

8a4422c934a1b33c8afec474734be684.png

If \(\hat{A} + \hat{C} = \text{180}°\) or \(\hat{B} + \hat{D} = \text{90}´\), then \(ABCD\) is a cyclic quadrilateral.
eb11576ecc960324046d75893cb27456.png

If \(\hat{A}_1 = \hat{C}\) or \(\hat{D}_1 = \hat{B}\), then \(ABCD\) is a cyclic quadrilateral.
25f0e3d06fbb3fb511a77e6ce01124aa.png

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

Tangents to a circle (EMCJ6)

cc5223443021195103ab80253111a402.png

A tangent is perpendicular to the radius (\(OT \perp ST\)), drawn to the point of contact with the circle.
d2436d5af7180a26dc8361aa073a6718.png

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)
8d86108694ce800d1b1f2b3c646a7de6.png
  • 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.

91b343e1387d3cac099f2ab8ddb74a2f.png

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

Revision

Textbook 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.

b8a873b56c75d0dadde1df96d2b90fc1.png
\[\begin{array}{rll} \hat{P} &= \frac{1}{2} M\hat{O}N & (\angle \text{ at centre = twice } \angle \text{ at circumference})\\ &= \text{30}° & \\ \therefore z &= \text{30}° &(\text{alt. } \angle\text{s, } MO \parallel NP) \end{array}\]

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

2f5fca65aac20d32e7ffea6970308318.png

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 \(P\hat{Q}S = a\).

133c45120483b7f08cba9055feb546fb.png

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 \(O\hat{C}D = \text{35}°\).

f6775d531dbb935c0c45295961df0d2a.png

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 \(F\).

c1b20d59b0f95475e139c12e2511d8a2.png

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 \(n\).

b9b7d244134babe67a91373a40731070.png
\[\begin{array}{rll} \hat{T} &= m &(PT = PR)\\ \therefore Q\hat{P}R &= \text{2}m &(\text{ext. } \angle \triangle = \text{ sum int. } \angle\text{s})\\ \therefore n &= \text{2}(\text{2}m)& (\angle\text{s at centre and circumference on } QR)\\ n &= 4m & \\ \therefore m &= \frac{1}{4}n & \end{array}\]

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

8747786a4a4aa29c38c52c015ede10d3.png
\[\begin{array}{rll} \text{In } &\triangle QOS, & \\ QP &= 30 & ( \text{given}) \\ QS &= \frac{1}{2}QP & (\perp \text{ from centre bisects chord}) \\ \therefore QS &= 15 & \\ QO^2 &= OS^2 + QS^2 & ( \text{Pythagoras}) \\ y^2 &= (y-9)^2 + 15^2 & \\ y^2 &= y^2 - 18y + 81 + 225 & \\ \therefore 18y &= 306 & \\ \therefore y &= \text{17}\text{ mm} & \end{array}\]

\(PQ\) 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.

6ad4e2d4408010b5ce7173756ceb3310.png

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 \(A\hat{T}B = x\).

7897a580a8c82275fc97faf092580446.png

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

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