What Grade 12 assumes you already know
Proved in Grade 11, accepted as given in Grade 12. If a row looks unfamiliar, revise it first.
| Theorem (given fact) | Condition | Reason to write | |
|---|---|---|---|
| Line from centre ⊥ to a chord bisects the chord (and the converse) | \(OM\perp AB\) | line from centre ⊥ chord | |
| The perpendicular bisector of a chord passes through the centre | \(AM=MB,\ OM\perp AB\) | perp bisector of chord | |
| ∠ at the centre = 2 × ∠ at the circumference (same arc) | \(\hat{O}=2\hat{C}\) on arc \(AB\) | ∠ at centre = 2∠ at circumference | |
| ∠s subtended by the same chord, on the same side, are equal | \(\hat{C_1}=\hat{C_2}\) subtend \(AB\) | ∠s in the same seg. | |
| ∠ subtended by a diameter (∠ in a semi-circle) = 90° | \(AB\) is a diameter | ∠ in semi-circle | |
| Opposite angles of a cyclic quadrilateral are supplementary | \(ABCD\) cyclic | opp ∠s of cyclic quad | |
| The exterior angle of a cyclic quadrilateral = the interior opposite angle | \(ABCD\) cyclic, \(DCE\) straight | ext ∠ of cyclic quad | |
| A tangent is perpendicular to the radius drawn to the point of contact | \(PT\) tangent at \(T\) | tan ⊥ radius | |
| Two tangents from the same external point are equal in length | \(PA,PB\) tangents from \(P\) | Tans from common pt. | |
| Tangent–chord angle = angle in the alternate segment | \(PT\) tangent, chord \(TA\) | tan chord theorem |
Worked revision proof — cyclic quadrilateral
Grade 11 · still examinableProve that the opposite angles of a cyclic quadrilateral are supplementary. (This exact instruction is a real, recurring NSC Paper 2 question — proving a Grade 11 result inside a Grade 12 paper.)
- Join \(OB\) and \(OD\).
- Reflex \(\hat{O}_{BD}=2\hat{A}\) (∠ at centre = 2∠ at circumference, both on arc \(BCD\))
- Non-reflex \(\hat{O}_{BD}=2\hat{C}\) (∠ at centre = 2∠ at circumference, both on arc \(BAD\))
- Reflex \(\hat{O}_{BD}+\) non-reflex \(\hat{O}_{BD}=360°\) (∠s around a point)
- \(2\hat{A}+2\hat{C}=360°\ \Rightarrow\ \hat{A}+\hat{C}=180°\) \(\blacksquare\)
In the diagram above, if reflex \(\hat{O}_{BD}=250°\), what is \(\hat{A}\)? And if \(\hat{C}=68°\), what is \(\hat{A}\)?
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How to think through any geometry proof
The four theorems Grade 12 actually adds
Examinable proofs — converses are not examinable.
Theorem 1 — Proportionality Theorem
Examinable proofIf a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. In \(\triangle ABC\), if \(DE \parallel BC\) with \(D\) on \(AB\) and \(E\) on \(AC\), then \(\dfrac{AD}{DB}=\dfrac{AE}{EC}\).
Diagram not drawn to scale.
Construction: join \(DC\) and \(EB\).
- \(\dfrac{\text{Area }\triangle ADE}{\text{Area }\triangle DEB}=\dfrac{AD}{DB}\) (equal height from \(E\), bases on \(AB\))
- \(\dfrac{\text{Area }\triangle ADE}{\text{Area }\triangle DEC}=\dfrac{AE}{EC}\) (equal height from \(D\), bases on \(AC\))
- Area \(\triangle DEB=\) Area \(\triangle DEC\) (same base \(DE\); \(B,C\) equidistant from \(DE\) since \(DE\parallel BC\))
- \(\therefore\ \dfrac{AD}{DB}=\dfrac{AE}{EC}\) \(\blacksquare\)
⚠ Don't mix part:whole with part:part — \(\tfrac{AD}{AB}\neq\tfrac{AD}{DB}\). ★ Spot it: line ∥ one side of a △, or "Midpoint Theorem" (\(AD=DB\)).
In \(\triangle ABC\), \(DE\parallel BC\), \(AD=6\), \(DB=4\), \(AE=9\). Find \(EC\).
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In \(\triangle ABC\), \(DE\parallel BC\), \(AB=15\), \(AD=9\), \(AE=12\). Find \(AC\).
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Theorem 2 — Equiangular Triangles are Similar
Examinable proofIf \(\triangle ABC\) and \(\triangle DEF\) have \(\hat A=\hat D\), \(\hat B=\hat E\), \(\hat C=\hat F\), then \(\triangle ABC \small{|\!|\!|} \triangle DEF\), i.e. \(\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF}\).
Diagram not drawn to scale.
Construction: mark \(P\) on \(ED\) with \(EP=BA\), and \(Q\) on \(EF\) with \(EQ=BC\). Join \(PQ\).
- \(\triangle EPQ\equiv\triangle BAC\) (SAS: \(EP=BA\), \(\hat E=\hat B\), \(EQ=BC\))
- \(\therefore\ \hat{EPQ}=\hat{A}=\hat D\) (given, and corresp. \(\angle\)s of \(\equiv\) \(\triangle\)s)
- \(\therefore\ PQ\parallel DF\) (corresp. \(\angle\)s equal)
- \(\dfrac{EP}{ED}=\dfrac{EQ}{EF}\) (proportionality theorem, Thm 1)
- \(\therefore\ \dfrac{BA}{ED}=\dfrac{BC}{EF}\), and similarly \(\dfrac{AC}{DF}\) is shown equal to both. \(\blacksquare\)
⚠ Match equal angles first, then read off corresponding sides — \(A\!\leftrightarrow\!D,B\!\leftrightarrow\!E,C\!\leftrightarrow\!F\). ★ Spot it: two pairs of equal angles (AA) is enough.
\(\triangle ABC \small{|\!|\!|} \triangle DEF\) with \(AB=8\), \(DE=12\), \(BC=10\). Find \(EF\).
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\(\triangle ABC \small{|\!|\!|} \triangle DEF\) with \(AB=4\), \(BC=6\), \(AC=8\), and \(DE=6\). Find \(EF\) and \(DF\).
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Theorem 3 — Triangles with Sides in Proportion are Similar
Examinable proofIf \(\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF}\), then \(\triangle ABC \small{|\!|\!|} \triangle DEF\).
Diagram not drawn to scale.
Construction: mark \(P\) on \(ED\) with \(EP=BA\), and \(Q\) on \(EF\) with \(EQ=BC\). Join \(PQ\).
- \(\dfrac{EP}{ED}=\dfrac{BA}{ED}=\dfrac{EQ}{EF}\) (given ratio, and \(EP=BA\), \(EQ=BC\))
- \(\therefore\ PQ\parallel DF\) (converse of proportionality theorem)
- \(\hat{EPQ}=\hat D\), \(\hat{EQP}=\hat F\) (corresp. \(\angle\)s, \(PQ\parallel DF\))
- \(\therefore\ \triangle EPQ \small{|\!|\!|} \triangle EDF\) (equiangular, Thm 2), so \(\dfrac{PQ}{DF}=\dfrac{EP}{ED}=\dfrac{BA}{ED}=\dfrac{BC}{EF}\) (given) \(\Rightarrow PQ=BC\)
- \(\triangle EPQ\equiv\triangle BAC\) (SSS: \(EP=BA\), \(PQ=BC\), \(EQ=AC\))
- \(\therefore\ \hat B=\hat E,\ \hat C=\hat F\), and \(\hat A=\hat D\) (shared) \(\Rightarrow \triangle ABC \small{|\!|\!|} \triangle DEF\) \(\blacksquare\)
⚠ Check all three ratios, not just two — two equal ratios alone can still be non-similar. ★ Spot it: three side lengths given, no angles.
\(\triangle ABC\) has sides \(6,8,10\). \(\triangle DEF\) has sides \(9,12,15\) in matching order. Are they similar? Give a reason.
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\(\triangle ABC\) has sides \(5,6,8\). \(\triangle DEF\) has sides \(10,12,15\) in matching order. Are they similar? Give a reason.
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Theorem 4 — Pythagoras via Similar Triangles
Examinable proofIn \(\triangle ABC\) with \(\hat A=90°\) and \(AD\perp BC\) (\(D\) on \(BC\)), prove that \(BC^2=AB^2+AC^2\).
Diagram not drawn to scale.
- In \(\triangle ABD\) and \(\triangle CBA\): \(\hat{ADB}=\hat{BAC}=90°\); \(\hat B\) is common. (∠∠, equiangular — Thm 2)
- \(\therefore\ \triangle ABD \small{|\!|\!|} \triangle CBA \Rightarrow \dfrac{AB}{CB}=\dfrac{BD}{BA}\ \Rightarrow\ AB^2=CB\cdot BD\) …(1)
- In \(\triangle ACD\) and \(\triangle BCA\): \(\hat{ADC}=\hat{BAC}=90°\); \(\hat C\) is common. (∠∠, equiangular — Thm 2)
- \(\therefore\ \triangle ACD \small{|\!|\!|} \triangle BCA \Rightarrow \dfrac{AC}{BC}=\dfrac{CD}{CA}\ \Rightarrow\ AC^2=BC\cdot CD\) …(2)
- (1) + (2): \(AB^2+AC^2=BC\cdot BD+BC\cdot CD=BC(BD+CD)=BC\cdot BC=BC^2\) \(\blacksquare\)
⚠ \(D\) is not the midpoint of \(BC\) in general — only \(BD+DC=BC\) is guaranteed. ★ Spot it: altitude from the right angle to the hypotenuse.
Using step 2 above: if \(CB=25\) and \(BD=9\), find \(AB\).
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Using step 4 above (\(AC^2=BC\cdot CD\)): if \(BC=20\) and \(CD=5\), find \(AC\).
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From recognition to a full NSC rider
In \(\triangle PQR\), \(ST\parallel QR\) with \(S\) on \(PQ\) and \(T\) on \(PR\). Which theorem lets you write \(\dfrac{PS}{SQ}=\dfrac{PT}{TR}\)?
The Proportionality Theorem (Theorem 1) — a line parallel to one side of a triangle divides the other two sides proportionally.
Complete the missing reason. In \(\triangle ABD\) and \(\triangle CBA\): \(\hat{ADB}=\hat{BAC}=90°\); \(\hat B\) common. \(\therefore \triangle ABD \small{|\!|\!|} \triangle CBA\) (reason: ______ )
∠∠ (equiangular triangles are similar — Theorem 2).
\(O\) is the centre of a circle with chord \(AB\). \(P\) lies on the major arc, and \(\hat{AOB}=2x\) (non-reflex). A tangent to the circle at \(A\) meets chord \(AB\), forming tangent–chord angle \(\hat T\) on the side of the minor arc. Determine \(\hat P\) and \(\hat T\) in terms of \(x\).
Hint 1
\(\hat{AOB}\) and \(\hat P\) both stand on the same chord \(AB\) — start with ∠ at centre = 2∠ at circumference.
Hint 2
The tangent-chord angle equals the angle in the alternate segment — that's the segment containing \(P\).
\(\hat P=x\) (∠ at centre = 2∠ at circumference: \(\hat{AOB}=2x\) and \(\hat{AOB}=2\hat P\) both subtend chord \(AB\), so \(2x=2\hat P \Rightarrow \hat P=x\)). Then \(\hat T=\hat P=x\) (tan chord theorem: the tangent–chord angle equals the angle in the alternate segment, which is exactly \(\hat P\)).
In \(\triangle DEF\), \(G\) is on \(DE\) and \(H\) is on \(DF\) such that \(DG=8\), \(GE=4\), \(DH=10\), \(HF=5\). Prove that \(GH\parallel EF\).
Hint
You aren't given that \(GH\parallel EF\) — you must prove it. Which theorem's converse connects two known ratios to a parallel conclusion?
\(\dfrac{DG}{GE}=\dfrac{8}{4}=2\) and \(\dfrac{DH}{HF}=\dfrac{10}{5}=2\). Since \(\dfrac{DG}{GE}=\dfrac{DH}{HF}\), \(GH\parallel EF\) (converse of the proportionality theorem — line divides two sides of a triangle in the same ratio).
In the diagram, \(AB\) is a diameter of a circle with centre \(O\). \(C\) is a point on the circle, and \(D\) is a point on \(AB\) such that \(CD\perp AB\).
- Write down the size of \(\hat{ACB}\), with a reason. (2)
- Prove that \(AC^2=AD\cdot AB\). (4)
- Hence, or otherwise, prove that \(CD^2=AD\cdot DB\). (4)
Hint for (1)
\(AB\) is a diameter — which Grade 11 circle theorem gives you the angle at \(C\) immediately?
Hint for (2)
Once \(\hat{ACB}=90°\) and \(CD\perp AB\), you have exactly the altitude-to-hypotenuse configuration from Theorem 4 above — just relabelled. Find two triangles that are similar for the same reason as in that proof.
Hint for (3)
Compare \(\triangle ACD\) and \(\triangle CBD\) this time, not \(\triangle ACD\) and the big triangle — both have a right angle at \(D\), so you only need one more equal angle.
(1) \(\hat{ACB}=90°\) (∠ in semi-circle — \(AB\) is a diameter).
(2) In \(\triangle ACD\) and \(\triangle ABC\): \(\hat{ADC}=\hat{ACB}=90°\); \(\hat A\) is common. \(\therefore \triangle ACD \small{|\!|\!|} \triangle ABC\) (∠∠). \(\therefore \dfrac{AC}{AB}=\dfrac{AD}{AC} \Rightarrow AC^2=AD\cdot AB\). \(\blacksquare\)
(3) In \(\triangle ACD\) and \(\triangle CBD\): \(\hat{ADC}=\hat{CDB}=90°\) (given \(CD\perp AB\)); \(\hat{ACD}=90°-\hat A\) (∠ sum of \(\triangle ACD\)) \(=\hat{ABC}\) (∠ sum of \(\triangle ABC\), since \(\hat{ACB}=90°\)) \(=\hat{DBC}\) (same angle). \(\therefore \triangle ACD \small{|\!|\!|} \triangle CBD\) (∠∠). \(\therefore \dfrac{AD}{CD}=\dfrac{CD}{DB} \Rightarrow CD^2=AD\cdot DB\). \(\blacksquare\)