Mathematics Grade 12 Euclidean Geometry Pdf

8.1 Revision (EMCHY)

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Types of triangles (EMCHZ)

Name	Diagram	Properties
Scalene		All sides and angles are different.
Isosceles		Two sides are equal in length. The angles opposite the equal sides are also equal.
Equilateral		All three sides are equal in length and all three angles are equal.
Acute-angled		Each of the three interior angles is less than \(\text{90}\)°.
Obtuse-angled		One interior angle is greater than \(\text{90}\)°.
Right-angled		One interior angle is \(\text{90}\)°.

Congruent triangles (EMCJ2)

Condition

Diagram

SSS

(side, side, side)

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

SAS

(side, incl. angle, side)

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

AAS

(angle, angle, side)

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

RHS

(\(\text{90}\)°, hypotenuse, side)

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

Similar triangles (EMCJ3)

Condition

Diagram

AAA

(angle, angle, angle)

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

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

SSS

(sides in prop.)

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

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

Circle geometry (EMCJ4)

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 \(O\).

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.

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

Cyclic quadrilaterals (EMCJ5)

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:

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

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

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

Tangents to a circle (EMCJ6)

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

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)

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.

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.

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

\(OD\)

\[\begin{array}{rll} \text{In } \triangle ODC, \quad OC^2 &= OD^2 + DC^2 &(\text{Pythagoras})\\ 5^2 &= OD^2 + 4^2 & \\ \therefore OD &= \text{3}\text{ cm} & \end{array}\]

\(AD\)

\[\begin{array}{rll} AO &= \text{5}\text{ cm} &\text{(radius)} \\ AD &= AO + OD & \\ &= \text{5} + \text{3} & \\ \therefore AD &= \text{8}\text{ cm} & \\ \end{array}\]

\(AB\)

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

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 \(PR\).

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

\(O\hat{D}C\)

\[\begin{array}{rll} OC &= OD &(\text{equal radii }) \\ \therefore O\hat{D}C &= \text{35}°&(\text{isosceles } \triangle OCD) \end{array}\]

\(C\hat{O}D\)

\[\begin{array}{rll} C\hat{O}D &= \text{180}° - \left( \text{35}° + \text{35}° \right)&(\text{sum } \angle\text{s } \triangle = \text{180}°) \\ &=\text{110}° & \end{array}\]

\(C\hat{B}D\)

\[\begin{array}{rll} C\hat{B}D &= \frac{1}{2} C\hat{O}D&(\angle\text{ at centre } = 2\angle\text{ at circum. } )\\ &= \text{55}° & \end{array}\]

\(B\hat{A}D\)

\[\begin{array}{rll} B\hat{A}D &= \text{90}° & (\angle\text{ in semi-circle}) \end{array}\]

\(A\hat{D}B\)

\[\begin{array}{rll} A\hat{D}B &= A\hat{B}D &(\text{isosceles } \triangle ABD) \\ \therefore A\hat{D}B &= \frac{\text{180}°-\text{90}°}{2}&(\text{sum } \angle\text{s in } \triangle = \text{180}°)\\ &= \text{45}° & \end{array}\]

\(C\hat{B}P = D\hat{B}P\)

\[\begin{array}{rll} \text{In } \triangle CBP &\text{ and } \triangle DBP\text{:}& \\ CP &= DP& (OP \perp CD)\\ C\hat{P}B &= D\hat{P}B = \text{90}° & (\text{given}) \\ BP &= BP & (\text{common})\\ \therefore \triangle CBP &\equiv \triangle DBP &(\text{SAS} ) \\ \therefore C\hat{B}P &= D\hat{B}P & (\triangle CBP \equiv \triangle DBP) \end{array}\\\]

\(C\hat{E}D = 2 C\hat{B}A\)

\[\begin{array}{rll} C\hat{E}D &= C\hat{B}D& (\angle\text{s on chord } CD)\\ \text{But } C\hat{B}A &= D\hat{B}A & (\triangle CBP \equiv \triangle DBP)\\ \therefore C\hat{E}D &= 2 C\hat{B}A \end{array}\]

\(A\hat{B}D = \frac{1}{2} C\hat{O}A\)

\[\begin{array}{rll} D\hat{B}A &= C\hat{B}A & (\triangle CBP \equiv \triangle DBP)\\ C\hat{B}A &= \frac{1}{2} C\hat{O}A &(\angle\text{ at centre } = 2\angle\text{ at circum. } )\\ \therefore A\hat{B}D &= \frac{1}{2} C\hat{O}A & \end{array}\]

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

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

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

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

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

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

\(BAPC\) is a cyclic quadrilateral

\[\begin{array}{rll} P\hat{C}Q &= B\hat{A}Q & ( \text{proven} ) \\ \therefore &BAPC \text{ is a cyclic quad. } & ( \text{ext. angle = int. opp. } \angle ) \end{array}\]

\(AB = AC\)

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

\(A\hat{B}T\)

\[\begin{array}{rll} A\hat{B}T &= B\hat{A}T & (TA = TB) \\ &= \frac{\text{180}° - x}{2} & ( \text{sum } \angle \text{s of } \triangle TAB) \\ &= \text{90}° - \frac{x}{2} \end{array}\]

\(O\hat{B}A\)

\[\begin{array}{rll} O\hat{B}T &= \text{90}° & (\text{tangent } \perp \text{ radius}) \\ \therefore O\hat{B}A &= \text{90}° - \left( \text{90}° - \frac{x}{2} \right) & \\ &= \frac{x}{2} \end{array}\]

\(\hat{C}\)

\[\begin{array}{rll} \hat{C} &= A\hat{B}T & (\text{tangent chord}) \\ &= \text{90}° - \frac{x}{2} & \end{array}\]