Euclidean Geometry Past Question Papers

How to use this bank.

Start at Level 1 and move up — don't jump to Level 4 first.
Draw the diagram and mark everything given before you touch a theorem.
Write the reason for every line, using the wording from the Summary Notes reference table.
Reveal the memo only after a genuine attempt.

Accuracy note: every question below was independently solved before publication. Source labels are exact — a specific citation means the question was checked against that real archived paper; anything not verifiable against the archive is labelled "Equation Station SA Practice Question," not a fabricated citation.

L1 — Knowledge 4 Qs

L2 — Routine Procedures 5 Qs

L3 — Complex Procedures 4 Qs

L4 — Problem Solving 4 Qs

20%

Level 1 | Knowledge

Recall & Direct Application

One theorem, one step. If these feel shaky, go back to the Summary Notes reference table before continuing.

Q1Equation Station SA Practice Question3 marks

Proportionality

Ratio from a Parallel Line

In \(\triangle PQR\), \(ST\parallel QR\) with \(S\) on \(PQ\) and \(T\) on \(PR\). \(PS=5\), \(SQ=3\), \(PT=8\). Calculate \(TR\).

Memo

✓ \(\dfrac{PS}{SQ}=\dfrac{PT}{TR}\) (line ∥ to one side of \(\triangle\))✓ \(\dfrac{5}{3}=\dfrac{8}{TR}\)✓ \(TR=\dfrac{8\times3}{5}=4{,}8\)

Q2Equation Station SA Practice Question2 marks

Circle theorems

Angle at the Centre

\(O\) is the centre of a circle. \(\hat{AOB}=120°\) (non-reflex), and \(C\) lies on the major arc. Calculate \(\hat{ACB}\).

Memo

✓ \(\hat{ACB}=\tfrac12\hat{AOB}=\tfrac12(120°)=60°\) (∠ at centre = 2∠ at circumference)

Q3Equation Station SA Practice Question2 marks

Cyclic quadrilateral

Opposite Angles

\(ABCD\) is a cyclic quadrilateral with \(\hat A=105°\). Calculate \(\hat C\).

Memo

✓ \(\hat A+\hat C=180°\) (opp ∠s of cyclic quad)✓ \(\hat C=180°-105°=75°\)

Q4Equation Station SA Practice Question4 marks

Tangents

Tangent Quadrilateral

\(PA\) and \(PB\) are tangents from external point \(P\) to a circle with centre \(O\), touching at \(A\) and \(B\). \(\hat{APB}=40°\). Calculate \(\hat{AOB}\).

Memo

✓ \(\hat{OAP}=\hat{OBP}=90°\) (tan ⊥ radius)✓ ∠s of quad \(OAPB\) sum to \(360°\)✓ \(\hat{AOB}=360°-90°-90°-40°=140°\)

35%

Level 2 | Routine Procedures

Apply a Named Theorem

Still one main theorem, but now with a full calculation or a short bookwork proof attached. The biggest mark bucket in the paper.

Q5Equation Station SA Practice Question3 marks

Midpoint Theorem

Midpoints of Two Sides

In \(\triangle ABC\), \(D\) and \(E\) are the midpoints of \(AB\) and \(AC\). \(AB=14\), \(AC=10\), \(BC=18\). Calculate \(DE\), naming the theorem used.

Memo

✓ \(DE=\tfrac12 BC\) (Midpoint Theorem — special case of the proportionality theorem)✓ \(DE=\tfrac12(18)=9\)

Q6Equation Station SA Practice Question4 marks

Similar triangles

Scale Factor from Similarity

\(\triangle ABC \small{|\!|\!|} \triangle DEF\). \(AB=6\), \(BC=9\), \(AC=12\), \(DE=8\). Calculate \(EF\) and \(DF\).

Memo

✓ scale factor \(=\dfrac{DE}{AB}=\dfrac{8}{6}=\dfrac43\)✓ \(EF=9\times\dfrac43=12\)✓ \(DF=12\times\dfrac43=16\)

Q7Equation Station SA Practice Question6 marks

Pythagoras (similar △)

Altitude to the Hypotenuse

In \(\triangle ABC\), \(\hat A=90°\) and \(AD\perp BC\), with \(D\) on \(BC\). \(BD=4\), \(DC=9\). Calculate \(AB\), \(AC\) and \(AD\).

Memo

✓ \(BC=BD+DC=13\)✓ \(AB^2=BC\cdot BD=13(4)=52\Rightarrow AB=\sqrt{52}=2\sqrt{13}\)✓ \(AC^2=BC\cdot DC=13(9)=117\Rightarrow AC=\sqrt{117}=3\sqrt{13}\)✓ \(AD^2=BD\cdot DC=4(9)=36\Rightarrow AD=6\) (altitude–geometric-mean relation, from \(\triangle ABD\small{|\!|\!|}\triangle CAD\))

Q8Equation Station SA Practice Question2 marks

Tangent-chord

Tangent–Chord Angle

\(PT\) is a tangent at \(T\), and \(TA\) is a chord with \(\hat{PTA}=55°\). Calculate the size of the inscribed angle in the alternate segment that subtends \(TA\).

Memo

✓ \(55°\) (tan chord theorem: tangent–chord ∠ = ∠ in alternate segment)

30%

Level 3 | Complex Procedures

Prove It, or Solve for an Unknown

A full bookwork proof, or a calculation that needs an equation set up first — not just a formula plugged in.

Q9Equation Station SA Practice Question4 marks

ProportionalitySolve for x

Unknown from a Ratio Equation

In \(\triangle ABC\), \(DE\parallel BC\) with \(D\) on \(AB\), \(E\) on \(AC\). \(AD=x\), \(DB=x+2\), \(AE=x+1\), \(EC=x+4\). Determine \(x\).

Memo

✓ \(\dfrac{x}{x+2}=\dfrac{x+1}{x+4}\) (line ∥ to one side of \(\triangle\))✓ \(x(x+4)=(x+2)(x+1)\)✓ \(x^2+4x=x^2+3x+2\)✓ \(x=2\)

Q10Equation Station SA Practice Question5 marks

TangentsProve

Prove: Two Tangents are Equal

\(PA\) and \(PB\) are tangents from an external point \(P\) to a circle with centre \(O\), touching at \(A\) and \(B\). Prove that \(PA=PB\).

Memo

✓ Join \(OA\), \(OB\), \(OP\).✓ In \(\triangle OAP\) and \(\triangle OBP\): \(OA=OB\) (radii); \(\hat{OAP}=\hat{OBP}=90°\) (tan ⊥ radius); \(OP\) common.✓ \(\therefore\triangle OAP\equiv\triangle OBP\) (RHS)✓ \(\therefore PA=PB\) \(\blacksquare\)

Q11Equation Station SA Practice Question4 marks

Similar trianglesProve

Prove Similarity from Parallel Lines

\(AB\parallel CD\). Diagonals \(AC\) and \(BD\) intersect at \(E\). Prove that \(\triangle AEB \small{|\!|\!|} \triangle CED\).

Memo

✓ \(\hat{AEB}=\hat{CED}\) (vert. opp. ∠s)✓ \(\hat{EAB}=\hat{ECD}\) (alt. ∠s, \(AB\parallel CD\))✓ \(\therefore\triangle AEB \small{|\!|\!|} \triangle CED\) (∠∠, equiangular \(\triangle\)s) \(\blacksquare\)

Q12Equation Station SA Practice Question4 marks

Midpoint Theorem

Perimeter of the Midpoint Triangle

In \(\triangle ABC\), \(D\), \(E\), \(F\) are the midpoints of \(AB\), \(AC\), \(BC\) respectively. The perimeter of \(\triangle ABC\) is \(40\) cm. Determine the perimeter of \(\triangle DEF\).

Memo

✓ \(DE=\tfrac12BC\), \(EF=\tfrac12AB\), \(DF=\tfrac12AC\) (Midpoint Theorem, applied three times)✓ Perimeter \(\triangle DEF=\tfrac12(AB+BC+AC)=\tfrac12(40)=20\) cm

15%

Level 4 | Problem Solving

Combine Theorems in an Unfamiliar Way

These riders don't announce which theorem to use — that's the point. Run the six-step routine from the Summary Notes page.

Q13Equation Station SA Practice Question10 marks

CircleSimilar △sIntegrated

Diameter, Altitude, and Two Length Relations

\(AB\) is a diameter of a circle centre \(O\). \(C\) is on the circle, and \(D\) is on \(AB\) with \(CD\perp AB\).

Write down \(\hat{ACB}\), with a reason. (2)
Prove that \(AC^2=AD\cdot AB\). (4)
Hence prove that \(CD^2=AD\cdot DB\). (4)

Memo

✓ (1) \(\hat{ACB}=90°\) (∠ in semi-circle) ✓ (2) In \(\triangle ACD,\triangle ABC\): \(\hat{ADC}=\hat{ACB}=90°\); \(\hat A\) common \(\Rightarrow\triangle ACD\small{|\!|\!|}\triangle ABC\) (∠∠) ✓ \(\dfrac{AC}{AB}=\dfrac{AD}{AC}\Rightarrow AC^2=AD\cdot AB\ \blacksquare\) ✓ (3) In \(\triangle ACD,\triangle CBD\): \(\hat{ADC}=\hat{CDB}=90°\); \(\hat{ACD}=90°-\hat A=\hat{ABC}=\hat{DBC}\) (∠ sums) \(\Rightarrow\triangle ACD\small{|\!|\!|}\triangle CBD\) ✓ \(\dfrac{AD}{CD}=\dfrac{CD}{DB}\Rightarrow CD^2=AD\cdot DB\ \blacksquare\)

Q14Equation Station SA Practice Question6 marks

TangentPythagoras

Tangent Length from an External Point

\(P\) is external to a circle with centre \(O\) and radius \(r=5\). \(PA\) is a tangent at \(A\), and \(PO=13\).

Prove that \(PA^2=PO^2-r^2\). (3)
Calculate \(PA\). (3)

Memo

✓ \(\hat{OAP}=90°\) (tan ⊥ radius), so \(\triangle OAP\) is right-angled at \(A\)✓ \(OP^2=OA^2+PA^2\) (Pythagoras) \(\Rightarrow PA^2=OP^2-OA^2=PO^2-r^2\ \blacksquare\)✓ \(PA^2=13^2-5^2=169-25=144\Rightarrow PA=12\)

Q15Equation Station SA Practice Question6 marks

CircleSimilar △sIntersecting chords

Intersecting Chords

Chords \(AB\) and \(CD\) of a circle intersect at \(E\), inside the circle. Prove that \(\triangle AEC \small{|\!|\!|} \triangle DEB\), and hence prove that \(AE\cdot EB=CE\cdot ED\).

Memo

✓ \(\hat{AEC}=\hat{DEB}\) (vert. opp. ∠s)✓ \(\hat{CAE}=\hat{CAB}=\hat{CDB}=\hat{BDE}\) (∠s in the same segment, both subtend chord \(CB\))✓ \(\therefore\triangle AEC\small{|\!|\!|}\triangle DEB\) (∠∠)✓ \(\dfrac{AE}{DE}=\dfrac{CE}{BE}\Rightarrow AE\cdot BE=CE\cdot DE\ \blacksquare\)

Q16KZN Pre-Paper 2, June 20255 marks

Cyclic quadrilateralProve

Prove the Cyclic Quadrilateral Theorem

\(JKLM\) is a cyclic quadrilateral with centre \(O\). Prove the theorem which states that \(\hat J+\hat L=180°\).

Memo

✓ Join \(OK\) and \(OM\).✓ Reflex \(\hat{O}_{KM}=2\hat J\) (∠ at centre = 2∠ at circumference, arc \(KLM\))✓ Non-reflex \(\hat{O}_{KM}=2\hat L\) (∠ at centre = 2∠ at circumference, arc \(KJM\))✓ Reflex \(\hat{O}_{KM}+\) non-reflex \(\hat{O}_{KM}=360°\) (∠s round a point)✓ \(2\hat J+2\hat L=360°\Rightarrow\hat J+\hat L=180°\ \blacksquare\)

Q17Equation Station SA Practice Question8 marks

CircleProportionalityIntegrated

Tangent, Chord and a Ratio

\(PT\) is a tangent to a circle at \(T\). A line through \(P\) cuts the circle at \(A\) and \(B\) (with \(A\) between \(P\) and \(B\)). Given that \(\triangle PTA \small{|\!|\!|} \triangle PBT\) (accept this as given), and \(PA=4\), \(AB=12\), determine \(PT\).

Memo

✓ \(PB=PA+AB=4+12=16\)✓ \(\triangle PTA\small{|\!|\!|}\triangle PBT\Rightarrow\dfrac{PT}{PB}=\dfrac{PA}{PT}\Rightarrow PT^2=PA\cdot PB\)✓ \(PT^2=4\times16=64\Rightarrow PT=8\)