Euclidean geometry — Mathematics Grade 12

9 exam-style question sets on Euclidean geometry (Paper 2), each with a hint and a fully worked answer. The app holds 45 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

Circle theorems: proof, cyclic quad in x & chord lengthsProportion theorem & similar trianglesTangents & cyclic quads: converse proofsCyclic quadrilateral: exterior angle & equal chordsProportionality theorem & midpoint theoremTangent circles: tan-chord, similar triangles & productsAngle at centre theorem: proof & tangent riderCyclic quadrilateral: perpendicular chord, parallel chords & diameterProportionality, similarity & tangents

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Question 1

Circle theorems: proof, cyclic quad in x & chord lengthsComplex16 marksPaper 2

This question uses THREE separate figures. No diagrams are supplied, so draw a neat sketch of each figure from its description before you start. Reasons must be given for ALL statements in parts (a) to (d).

Figure for (a): $G$ , $H$ and $K$ are points on a circle with centre $O$ , placed so that $O$ lies inside $△ G H K$ . The radii $O G$ and $O H$ are drawn, and so are the chords $K G$ and $K H$ .

Figure for (b): $P QR S$ is a cyclic quadrilateral in a circle with centre $M$ , and $M$ lies inside $P QR S$ . The radii $M Q$ and $M S$ are drawn. The non-reflex angle subtended at $M$ by arc $QR S$ is $Q \hat{M} S = (2 x + 40)^{\circ}$ , and $Q \hat{R} S = (3 x - 16)^{\circ}$ .

Figure for (c) and (d): $N$ is the centre of a larger circle and $N A$ is a radius of this circle. A smaller circle is drawn with $N A$ as its diameter. The chord $A B$ of the larger circle cuts the smaller circle again at $C$ , and $N C$ is drawn.

aProve the theorem which states that the angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at the circumference, i.e. prove that $G \hat{O} H = 2 G \hat{K} H$ . (5)
bCalculate, with reasons, the value of $x$ . (5)
cWrite down, with a reason, the size of $N \hat{C} A$ . (2)
dGiven that $N A = 25$ cm and $A B = 48$ cm, calculate, with reasons, the length of $N C$ . (4)

Hint

(a) Join $K$ to $O$ and extend the line beyond $O$ : this creates two isosceles triangles with their exterior angles at $O$ . (b) First write $Q \hat{P} S$ in terms of $x$ , then recall how opposite angles of a cyclic quadrilateral are related. (c) and (d) $N A$ is a diameter of the smaller circle; once you know the angle at $C$ , decide where $C$ lies on the chord $A B$ before you use Pythagoras.

Worked answer

a) Construction: join $K O$ and produce it to a point $L$ beyond $O$ . ✓
In $△ GO K$ : $O G = O K$ (radii), so $O \hat{G} K = O \hat{K} G$ (∠s opp equal sides) ✓
$G \hat{O} L = O \hat{G} K + O \hat{K} G = 2 O \hat{K} G$ (ext ∠ of △) ✓
Similarly, in $△ H O K$ : $H \hat{O} L = 2 O \hat{K} H$ ✓
∴ $G \hat{O} H = G \hat{O} L + H \hat{O} L = 2 (O \hat{K} G + O \hat{K} H) = 2 G \hat{K} H$ ✓

b) $Q \hat{P} S = \frac{1}{2} Q \hat{M} S = \frac{1}{2} (2 x + 40) = x + 20$ ✓ (∠ at centre $= 2 \times$ ∠ at circumf.) ✓
$Q \hat{P} S + Q \hat{R} S = 18 0^{\circ}$ (opp ∠s of cyclic quad) ✓
$x + 20 + 3 x - 16 = 180$ ✓
$4 x = 176$ , so $x = 44$ ✓
(Check: $Q \hat{M} S = 12 8^{\circ}$ , $Q \hat{P} S = 6 4^{\circ}$ and $Q \hat{R} S = 11 6^{\circ}$ ; $6 4^{\circ} + 11 6^{\circ} = 18 0^{\circ}$ .)

c) $N A$ is a diameter of the smaller circle, so $N \hat{C} A = 9 0^{\circ}$ ✓ (∠ in semi-circle) ✓

d) $N C ⊥ A B$ from (c), so $A C = C B = \frac{1}{2} (48) = 24$ cm (line from centre ⊥ to chord) ✓
In $△ N C A$ : $N C^{2} = N A^{2} - A C^{2}$ (Pythagoras) ✓
$N C^{2} = 2 5^{2} - 2 4^{2} = 625 - 576 = 49$ ✓
$N C = 7$ cm ✓

Modelled on Nov 2023 Paper 2, Question 8 — same skills, new scenario.

Question 2

Proportion theorem & similar trianglesComplex9 marksPaper 2

No diagram is provided: sketch the figure yourself from the description below, which contains everything you need. Support ALL statements and calculations with reasons.

Figure: $W X Y Z$ is a parallelogram with $W X ∥ Z Y$ and $W Z ∥ X Y$ , and its diagonal $W Y$ is drawn. $J$ is the point on side $W Z$ for which $W J : J Z = 5 : 3$ . A line through $J$ , parallel to $W X$ , crosses the diagonal $W Y$ at $M$ and meets side $X Y$ at $N$ . It is given that $W X = 24$ cm and $W Y = 32$ cm.

aCalculate, with reasons, the length of $M Y$ . (3)
bProve that $△ Y M N ∣∣∣ △ Y W X$ . (3)
cHence, calculate the length of $J M$ . Give reasons. (3)

Hint

Because $W X ∥ Z Y$ , the line through $J$ is parallel to $Z Y$ as well: in $△ W Z Y$ the segment $J M$ therefore divides $W Y$ in the same ratio as $W J : J Z$ . For (b), the angle at $Y$ belongs to both triangles, and $M N ∥ W X$ gives corresponding angles. In (c), use the ratio $\frac{Y M}{Y W}$ to find $M N$ first; then notice that $W J N X$ is itself a parallelogram, which tells you the length of $J N$ .

Worked answer

a) $J N ∥ W X$ (given) and $W X ∥ Z Y$ (opp sides of ||m), so in $△ W Z Y$ : $J M ∥ Z Y$
$\frac{Y M}{M W} = \frac{Z J}{J W}$ ✓ (prop theorem; JM ∥ ZY) ✓ $= \frac{3}{5}$
$∴ M Y = \frac{3}{8} \times W Y = \frac{3}{8} \times 32 = 12$ cm ✓

b) In $△ Y M N$ and $△ Y W X$ :
$\hat{Y}$ is common ✓
$Y \hat{M} N = Y \hat{W} X$ (corresp ∠s; MN ∥ WX) ✓
$Y \hat{N} M = Y \hat{X} W$ (corresp ∠s; MN ∥ WX)
$∴ △ Y M N ∣∣∣ △ Y W X$ (∠∠∠) ✓

c) $\frac{M N}{W X} = \frac{Y M}{Y W}$ (|||△s) ✓ $= \frac{12}{32} = \frac{3}{8}$ , so $M N = \frac{3}{8} \times 24 = 9$ cm ✓
$W J ∥ X N$ (opp sides of ||m) and $J N ∥ W X$ (given), so $W J N X$ is a parallelogram (both pairs of opp sides ∥) and $J N = W X = 24$ cm (opp sides of ||m)
$∴ J M = J N - M N = 24 - 9 = 15$ cm ✓
(OR: $△ W J M ∣∣∣ △ W Z Y$ , so $\frac{J M}{Z Y} = \frac{W J}{W Z} = \frac{5}{8}$ ; with $Z Y = W X = 24$ cm (opp sides of ||m), $J M = 15$ cm.)

Modelled on Nov 2023 Paper 2, Question 9 — same skills, new scenario.

Question 3

Tangents & cyclic quads: converse proofsComplex15 marksPaper 2

No diagram is provided: the figure is described fully below, and the description is authoritative. Draw a neat sketch of it before you start. Give reasons for ALL statements.

Figure: $A$ , $B$ , $C$ and $D$ lie on a circle in that order, so that $A B C D$ is a cyclic quadrilateral in which $A B = A C$ . The tangents to the circle at $A$ and at $C$ meet at $T$ , and $T$ lies on $B D$ produced, with $D$ between $B$ and $T$ . The diagonals $A C$ and $B D$ intersect at $K$ . Side $A D$ is produced beyond $D$ to meet the tangent $T C$ at $G$ , which lies between $T$ and $C$ .

aProve that $T \hat{A} C = A \hat{D} B$ . (5)
bProve that $K D GC$ is a cyclic quadrilateral. (4)
cProve that $G K$ is a tangent to the circle that passes through $A$ , $K$ and $D$ . (6)

Hint

(a) Begin at the tangent $T A$ : the tan-chord theorem moves $T \hat{A} C$ into the circle, and the isosceles triangle $A B C$ turns it into an angle that is also subtended at $D$ . (b) The two tangents from $T$ are equal, so $△ T A C$ is isosceles; then look for an exterior angle of $K D GC$ at $D$ . (c) Use circle $K D GC$ to move $G \hat{K} D$ to $C$ , and finish with the tan-chord theorem at $C$ .

Worked answer

a) $T \hat{A} C = A \hat{B} C$ ✓ (tan-chord) ✓
$A \hat{B} C = A \hat{C} B$ (∠s opp equal sides; $A B = A C$ ) ✓
$A \hat{C} B = A \hat{D} B$ (∠s in same seg) ✓
$∴ T \hat{A} C = A \hat{D} B$ ✓

b) $T A = T C$ (tans from common pt), so $T \hat{C} A = T \hat{A} C$ (∠s opp equal sides) ✓
$T \hat{D} G = A \hat{D} B$ (vert opp ∠s) ✓
Using (a): $T \hat{D} G = A \hat{D} B = T \hat{A} C = T \hat{C} A = G \hat{C} K$ ( $G$ lies on $C T$ and $K$ lies on $C A$ ) ✓
$T \hat{D} G$ is the exterior angle of $K D GC$ at $D$ (because $K D$ produced passes through $T$ ), and $G \hat{C} K$ is the interior opposite angle, so $K D GC$ is a cyclic quad (ext ∠ = int opp ∠) ✓

c) $G \hat{K} D = G \hat{C} D$ ✓ (∠s in same seg; $K D GC$ is cyclic) ✓
$G \hat{C} D = T \hat{C} D$ ( $G$ lies on $C T$ ) and $T \hat{C} D = C \hat{A} D$ ✓ (tan-chord) ✓
$C \hat{A} D = K \hat{A} D$ ( $K$ lies on $A C$ ), so $G \hat{K} D = K \hat{A} D$ ✓
$∴ G K$ is a tangent to the circle through $A$ , $K$ and $D$ (converse tan-chord) ✓

Modelled on Nov 2023 Paper 2, Question 10 — same skills, new scenario.

Question 4

Cyclic quadrilateral: exterior angle & equal chordsComplex6 marksPaper 2

$A$ , $B$ , $C$ and $D$ lie on a circle in that order, so that $A B C D$ is a cyclic quadrilateral. Side $D C$ is produced to $E$ , and the diagonal $A C$ is drawn. The diagonal divides the angle at $A$ into $\hat{A}_{1} = D \hat{A} C$ and $\hat{A}_{2} = C \hat{A} B$ , while the angles at $C$ are numbered $\hat{C}_{1} = D \hat{C} A$ , $\hat{C}_{2} = A \hat{C} B$ and $\hat{C}_{3} = B \hat{C} E$ . It is given that $\hat{C}_{3} = 9 6^{\circ}$ , $A \hat{D} C = 12 8^{\circ}$ and $\hat{A}_{2} = 2 \hat{A}_{1}$ . Use this description to make your own neat sketch before answering. Reasons must be given for ALL statements.

aCalculate, giving reasons, the size of $\hat{A}_{2}$ . (2)
bProve that $A B = B C$ . (4)

Hint

Link the exterior angle at $C$ to the whole angle at $A$ first; then work inside $△ A B C$ until you find a second inscribed angle equal to $\hat{A}_{2}$ .

Worked answer

a) $\hat{A}_{1} + \hat{A}_{2} = \hat{C}_{3} = 9 6^{\circ}$ (ext ∠ of cyclic quad) ✓ With $\hat{A}_{2} = 2 \hat{A}_{1}$ : $3 \hat{A}_{1} = 9 6^{\circ}$ , so $\hat{A}_{1} = 3 2^{\circ}$ and $\hat{A}_{2} = 6 4^{\circ}$ ✓

b) $A \hat{B} C = 18 0^{\circ} - 12 8^{\circ} = 5 2^{\circ}$ (opp ∠s of cyclic quad) ✓
In $△ A B C$ : $\hat{C}_{2} = 18 0^{\circ} - 5 2^{\circ} - 6 4^{\circ} = 6 4^{\circ}$ (∠ sum of △) ✓
Now $\hat{C}_{2}$ is the ∠ that chord $A B$ subtends at $C$ , and $\hat{A}_{2}$ is the ∠ that chord $B C$ subtends at $A$ , with $\hat{C}_{2} = \hat{A}_{2} = 6 4^{\circ}$ ✓
∴ $A B = B C$ (equal ∠s; equal chords) ✓

Modelled on Nov 2024 Paper 2, Question 9 — same skills, new scenario.

Question 5

Proportionality theorem & midpoint theoremComplex15 marksPaper 2

Answer this question WITHOUT a diagram: each figure is described completely in words below, and the description serves as the figure. Reasons must be given for ALL geometry statements.

Figure for (a): In $△ K L M$ , point $P$ lies on side $K L$ and point $Q$ lies on side $K M$ such that $P Q ∥ L M$ . You may add any construction lines that your proof needs.

Figure for (b) and (c): $O$ is the centre of a circle. $A$ and $B$ lie on the circle, and the chord $A B$ is NOT a diameter. $N$ is the point on $A B$ for which $O N ⊥ A B$ . The line from $B$ through $O$ is produced to cut the circle again at $C$ , so that $B C$ is a diameter, and the chord $A C$ is drawn. $D$ is the point on $A C$ with $A D : D C = 1 : 3$ , and the line through $D$ parallel to $A B$ cuts $B C$ at $E$ .

aProve the theorem which states that $\frac{K P}{P L} = \frac{K Q}{QM}$ . (6)
bProve that $A C = 2 O N$ . (5)
cIf $B C = 24$ cm, calculate the length of $E B$ . (4)

Hint

In (a) compare areas of triangles that share a height; in (b) look for the midpoints of $B A$ and $B C$ and use the midpoint theorem; in (c) let $D E ∥ A B$ carry the ratio $3 : 1$ onto $B C$ .

Worked answer

a) Construction: join $L Q$ and $M P$ ; let $h_{1}$ be the perpendicular height from $Q$ to $K L$ and $h_{2}$ the perpendicular height from $P$ to $K M$ . ✓

$\frac{area △ K P Q}{area △ P L Q} = \frac{\frac{1}{2} \cdot K P \cdot h _{1}}{\frac{1}{2} \cdot P L \cdot h _{1}} = \frac{K P}{P L}$ (△s with same height) ✓

$\frac{area △ K P Q}{area △ QM P} = \frac{\frac{1}{2} \cdot K Q \cdot h _{2}}{\frac{1}{2} \cdot QM \cdot h _{2}} = \frac{K Q}{QM}$ (△s with same height) ✓

area $△ P L Q =$ area $△ QM P$ ✓ (same base $P Q$ ; between same ∥ lines $P Q$ and $L M$ ) ✓

$∴ \frac{area △ K P Q}{area △ P L Q} = \frac{area △ K P Q}{area △ QM P}$ , so $\frac{K P}{P L} = \frac{K Q}{QM}$ ✓

b) $A N = N B$ ✓ (line from centre ⊥ to chord) ✓
$B O = O C$ (radii), so $O$ is the midpoint of $B C$ ✓
In $△ B A C$ : $N$ and $O$ are the midpoints of sides $B A$ and $B C$ , so $O N = \frac{1}{2} A C$ ✓ (Midpt Theorem) ✓
$∴ A C = 2 O N$

c) In $△ C A B$ : $D E ∥ A B$ , so $\frac{C D}{D A} = \frac{C E}{E B}$ ✓ (prop theorem; DE ∥ AB) ✓
$A D : D C = 1 : 3$ , so $\frac{C E}{E B} = \frac{3}{1}$ and $E B = \frac{1}{4} \times B C$ ✓ $= \frac{1}{4} \times 24 = 6$ cm ✓

Modelled on Nov 2024 Paper 2, Question 10 — same skills, new scenario.

Question 6

Tangent circles: tan-chord, similar triangles & productsComplex20 marksPaper 2

Two circles touch each other internally at $A$ , the smaller circle lying inside the larger one. $S A T$ is the common tangent to both circles at $A$ . The chords $A B$ and $A D$ of the larger circle cut the smaller circle again at $C$ and $F$ respectively. The straight line through $A$ and the two centres separates the figure so that $T$ , $B$ and $C$ lie on one side of it, while $S$ , $D$ and $F$ lie on the other side. The chord $B D$ of the larger circle touches the smaller circle at $E$ , where $E$ lies between $B$ and $D$ . $A E$ is produced to meet the larger circle again at $G$ , and $C E$ , $C F$ and $B G$ are drawn. The angles at $E$ are numbered $\hat{E}_{1} = B \hat{E} C$ and $\hat{E}_{2} = C \hat{E} A$ . It is given that $T \hat{A} B = x$ . Provide a reason for EVERY statement in your answers.

aName, with reasons, FOUR other angles in the figure that are each equal to $x$ . (6)
bProve that $A E \cdot A B = A G \cdot A C$ . (4)
cProve that $△ B E C$ ||| $△ B A E$ and hence that $B E^{2} = B C \cdot B A$ . (4)
dUse the results of parts (b) and (c) to prove that $A B^{2} \cdot E G = B E^{2} \cdot A G$ . (6)

Hint

Chase each angle back to the tangent at $A$ with the tan-chord theorem, prove both similarities from the equal angles of part (a), and finish by writing $B C$ as $A B - A C$ before substituting.

Worked answer

a) $A \hat{D} B = x$ ✓ (tan-chord — tangent $T A$ , chord $A B$ of the larger circle) ✓
$A \hat{F} C = x$ ✓ (tan-chord — tangent $T A$ , chord $A C$ of the smaller circle) ✓
$A \hat{E} C = A \hat{F} C = x$ ✓ (∠s in same seg — $E$ and $F$ subtend chord $A C$ in the smaller circle)
$A \hat{G} B = A \hat{D} B = x$ ✓ (∠s in same seg — $G$ and $D$ subtend chord $A B$ in the larger circle)

b) In △ $A E C$ and △ $A GB$ :
$E \hat{A} C = G \hat{A} B$ (common ∠ — $C$ lies on $A B$ and $E$ lies on $A G$ ) ✓
$A \hat{E} C = A \hat{G} B = x$ (proved in (a)) ✓
$∴$ △ $A E C$ ||| △ $A GB$ (AA) ✓
$∴ \frac{A E}{A G} = \frac{E C}{GB} = \frac{A C}{A B}$ , and cross-multiplying: $A E \cdot A B = A G \cdot A C$ ✓

c) In △ $B E C$ and △ $B A E$ :
$\hat{B}$ is common ✓
$\hat{E}_{1} = B \hat{E} C = C \hat{A} E$ (tan-chord — $B D$ touches the smaller circle at $E$ , chord $E C$ ); since $C$ lies on $A B$ , $C \hat{A} E = B \hat{A} E$ , so $B \hat{E} C = B \hat{A} E$ ✓
$∴$ △ $B E C$ ||| △ $B A E$ (AA) ✓
$∴ \frac{B E}{B A} = \frac{E C}{A E} = \frac{B C}{B E}$ , so $B E^{2} = B C \cdot B A$ ✓

d) From (b): $A C = \frac{A E \cdot A B}{A G}$ ✓
$C$ lies between $A$ and $B$ , so $B C = A B - A C = A B - \frac{A E \cdot A B}{A G} = \frac{A B ( A G - A E )}{A G}$ ✓✓
$E$ lies between $A$ and $G$ , so $A G - A E = E G$ and $B C = \frac{A B \cdot E G}{A G}$ ✓
Substitute into (c): $B E^{2} = B C \cdot B A = \frac{A B \cdot E G}{A G} \cdot A B = \frac{A B ^{2} \cdot E G}{A G}$ ✓
$∴ A B^{2} \cdot E G = B E^{2} \cdot A G$ ✓

Modelled on Nov 2024 Paper 2, Question 11 — same skills, new scenario.

Question 7

Angle at centre theorem: proof & tangent riderKnowledge10 marksPaper 2

In the first figure, $A$ , $B$ and $C$ are points on the circumference of a circle with centre $O$ , placed so that $O$ lies inside $△ A B C$ . The radii $O A$ and $O B$ are drawn, as well as the chords $C A$ and $C B$ . In the second figure, $K L$ is a diameter of a DIFFERENT circle with centre $M$ , and the tangent to this circle at $L$ is drawn. $N$ is a point on this second circle, above the diameter $K L$ , such that $N \hat{K} L = 3 4^{\circ}$ , and $M N$ produced meets the tangent at $T$ . Use these descriptions to make your own neat sketch of each figure before answering. Give reasons for your statements in parts (b) and (c).

aProve the theorem which states that the angle which arc $A B$ subtends at the centre $O$ is TWICE the angle that this arc subtends at $C$ on the circumference, i.e. prove that $A \hat{O} B = 2 A \hat{C} B$ . (5)
bCalculate, giving a reason, the size of $N \hat{M} L$ . (2)
cCalculate, giving reasons, the size of $M \hat{T} L$ . (3)

Hint

For the proof, join $C O$ and extend it beyond $O$ , then use two isosceles radius triangles; in the rider, part (a)'s theorem gives the angle at $M$ , and radius $M L$ meets the tangent at $9 0^{\circ}$ .

Worked answer

a) Construction: join $C O$ and produce it to a point $D$ beyond $O$ . ✓
In $△ A O C$ : $O A = O C$ (radii), so $O \hat{A} C = O \hat{C} A$ (∠s opp equal sides) ✓
$A \hat{O} D = O \hat{A} C + O \hat{C} A = 2 O \hat{C} A$ (ext ∠ of △) ✓
Similarly, in $△ B O C$ : $B \hat{O} D = 2 O \hat{C} B$ ✓
∴ $A \hat{O} B = A \hat{O} D + B \hat{O} D = 2 (O \hat{C} A + O \hat{C} B) = 2 A \hat{C} B$ ✓

b) $N \hat{M} L = 2 \times N \hat{K} L$ (∠ at centre $= 2 \times$ ∠ at circumf.) ✓ $= 2 \times 3 4^{\circ} = 6 8^{\circ}$ ✓

c) $M \hat{L} T = 9 0^{\circ}$ (tan ⊥ radius) ✓
$T$ lies on $M N$ produced, so $L \hat{M} T = N \hat{M} L = 6 8^{\circ}$ ; in $△ M L T$ : $M \hat{T} L = 18 0^{\circ} - 9 0^{\circ} - 6 8^{\circ}$ (∠ sum of △) ✓
$M \hat{T} L = 2 2^{\circ}$ ✓

Modelled on Nov 2025 Paper 2, Question 9 — same skills, new scenario.

Question 8

Cyclic quadrilateral: perpendicular chord, parallel chords & diameterComplex8 marksPaper 2

$K$ , $T$ , $L$ , $M$ and $N$ are five points lying in that order on a circle, so that $K L M N$ is a cyclic quadrilateral with $T$ on arc $K L$ . The chord drawn from $N$ cuts the side $K L$ perpendicularly at the point $F$ between $K$ and $L$ , and meets the circle again at $T$ . The chord $K T$ and the diagonal $N L$ are also drawn. The angles are labelled $\hat{K}_{1} = T \hat{K} L$ , $\hat{L}_{1} = K \hat{L} N$ , $\hat{L}_{2} = N \hat{L} M$ , $\hat{N}_{1} = K \hat{N} T$ , $\hat{N}_{2} = T \hat{N} L$ and $\hat{N}_{3} = L \hat{N} M$ . It is given that $\hat{L}_{1} = 3 2^{\circ}$ and that $\hat{K}_{1} = \hat{L}_{2}$ . Use this description to make your own neat sketch before answering. Reasons must be given for ALL statements.

aCalculate, giving reasons, the size of $\hat{K}_{1}$ . (3)
bProve, giving reasons, that $N T ∥ M L$ . (3)
cProve, giving reasons, that $K M$ is a diameter of the circle. (2)

Hint

Close the right-angled $△ N F L$ first, remember that $\hat{N}_{2}$ and $\hat{K}_{1}$ stand on the same arc $T L$ , and for the last part look at the size of the whole angle $K \hat{L} M$ .

Worked answer

a) In $△ N F L$ : $N \hat{F} L = 9 0^{\circ}$ ( $N T ⊥ K L$ , given) ✓
$\hat{N}_{2} = 18 0^{\circ} - 9 0^{\circ} - 3 2^{\circ} = 5 8^{\circ}$ (∠ sum of △) ✓
$\hat{K}_{1} = \hat{N}_{2} = 5 8^{\circ}$ (∠s in same seg, both subtended by chord $T L$ ) ✓

b) $\hat{N}_{2} = \hat{K}_{1}$ (∠s in same seg) ✓
$\hat{K}_{1} = \hat{L}_{2}$ (given), so $\hat{N}_{2} = \hat{L}_{2}$ ✓
But $\hat{N}_{2}$ and $\hat{L}_{2}$ are alternate angles formed by $N T$ and $M L$ with transversal $N L$
∴ $N T ∥ M L$ (alt ∠s equal) ✓

c) $K \hat{L} M = \hat{L}_{1} + \hat{L}_{2} = 3 2^{\circ} + 5 8^{\circ} = 9 0^{\circ}$ , since $\hat{L}_{2} = \hat{K}_{1} = 5 8^{\circ}$ from (a) ✓
∴ $K M$ is a diameter (chord subtends 90°) ✓

Modelled on Nov 2025 Paper 2, Question 10 — same skills, new scenario.

Question 9

Proportionality, similarity & tangentsComplex23 marksPaper 2

This question has NO diagram: each configuration is specified completely in words below, and the description serves as the figure. Reasons must be given for ALL geometry statements.

Figure for (a) to (c): In $△ A B C$ , $D$ is a point on side $A B$ such that $A D : D B = 3 : 2$ . The line through $D$ parallel to $B C$ cuts $A C$ at $E$ . Side $B C$ is produced to $F$ , with $C$ between $B$ and $F$ , such that $B C : C F = 2 : 1$ . $A F$ is drawn, and $D E$ produced cuts $A F$ at $H$ .

Figure for (d) to (h): $K L N M$ is a cyclic quadrilateral in which $M K = M L$ . The tangent to the circle at $K$ meets $M L$ produced at $T$ , so that $L$ lies between $M$ and $T$ . From $T$ a second tangent is drawn, touching the circle at $N$ . The line through $T$ parallel to chord $K L$ meets $M K$ produced at $S$ , so that $K$ lies between $M$ and $S$ . It is given that $△ T K L ∣∣∣ △ T M K$ .

aDetermine, with reasons, the ratio $\frac{A E}{E C}$ . (2)
bDetermine, with reasons, the ratio $E H : B C$ . (4)
cDetermine the ratio of the area of $△ A D E$ to the area of quadrilateral $D B C E$ . (4)
dProve that $K S = L T$ . (2)
eWrite down $T K^{2}$ in terms of $T L$ and $T M$ . (1)
fProve that $△ S T K ∣∣∣ △ S M T$ . (3)
gDeduce that $S T = T K$ . (3)
hHence show that $T N = S K \cdot S M$ . (4)

Hint

Carry the ratio $3 : 2$ across each parallel line with the proportionality theorem, square the similarity ratio for the areas, and in the circle figure chain (d), (e) and (f) together so that the equal tangents from $T$ deliver part (h).

Worked answer

a) In $△ A B C$ : $D E ∥ B C$ , so $\frac{A E}{E C} = \frac{A D}{D B}$ ✓ (prop theorem; DE ∥ BC) $= \frac{3}{2}$ ✓

b) $H$ lies on $D E$ produced and $F$ on $B C$ produced, so $E H ∥ C F$ . In $△ A E H$ and $△ A C F$ : $\hat{A}$ is common and $∠ A E H = ∠ A C F$ (corresp ∠s; EH ∥ CF), so $△ A E H ∣∣∣ △ A C F$ ✓
$∴ \frac{E H}{C F} = \frac{A E}{A C} = \frac{3}{5}$ ✓ (from (a), $A E : E C = 3 : 2$ )
$C F = \frac{1}{2} B C$ ✓ (given $B C : C F = 2 : 1$ )
$∴ E H = \frac{3}{5} \times \frac{1}{2} B C = \frac{3}{10} B C$ , i.e. $E H : B C = 3 : 10$ ✓

c) In $△ A D E$ and $△ A B C$ : $\hat{A}$ is common and $∠ A D E = ∠ A B C$ (corresp ∠s; DE ∥ BC), so $△ A D E ∣∣∣ △ A B C$ ✓
$\frac{area △ A D E}{area △ A B C} = (\frac{A D}{A B})^{2}$ ✓ (|||△s) $= (\frac{3}{5})^{2} = \frac{9}{25}$ ✓
area quad $D B C E =$ area $△ A B C -$ area $△ A D E$ , i.e. $25 - 9 = 16$ parts
$∴ \frac{area △ A D E}{area quad D B C E} = \frac{9}{16}$ ✓

d) In $△ M S T$ : $K L ∥ S T$ , so $\frac{M K}{K S} = \frac{M L}{L T}$ ✓ (prop theorem; KL ∥ ST)
$M K = M L$ (given) $∴ K S = L T$ ✓

e) $T K^{2} = T L \cdot T M$ ✓ (corresp sides; given $△ T K L ∣∣∣ △ T M K$ )

f) In $△ S T K$ and $△ S M T$ :
$\hat{S}$ is common ( $K$ lies on $S M$ )
$∠ S T K = ∠ T K L$ ✓ (alt ∠s; ST ∥ KL) $= ∠ K M L$ ✓ (tan-chord) $= ∠ S M T$ ( $S$ on $M K$ produced, $T$ on $M L$ produced)
$∴ △ S T K ∣∣∣ △ S M T$ ✓ (AA)

g) From (f): $\frac{S T}{S M} = \frac{S K}{S T}$ , so $S T^{2} = S K \cdot S M$ ✓
$S K = L T$ (part (d)) and $S M = S K + K M = L T + L M = T M$ ✓ ( $M K = M L$ , given)
$∴ S T^{2} = L T \cdot T M = T K^{2}$ (part (e)), and since lengths are positive, $S T = T K$ ✓

h) $T N = T K$ ✓ (tans from common pt)
$T K = S T$ ✓ (part (g))
$S T^{2} = S K \cdot S M$ ✓ (part (f), as used in (g))
$∴ T N^{2} = S K \cdot S M$ , i.e. $T N = S K \cdot S M$ ✓

Modelled on Nov 2025 Paper 2, Question 11 — same skills, new scenario.

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