IEB · Grade 12 · Mathematics Paper 2 · Euclidean geometry
In the print dialog, choose Save as PDF. IEB Grade 12 Euclidean Geometry Cheat Sheet Euclidean geometry is about 40 marks of Paper 2. Every statement needs a reason, so this sheet gives the accepted short form for each. Circle theorems (reason in brackets) Line from centre perpendicular to a chord bisects it (line from centre $⊥$ chord). Line from centre to midpoint of a chord is perpendicular to it (line from centre to midpt of chord). Angle at the centre is twice the angle at the circumference ( $∠$ at centre $= 2 \times ∠$ at circumference). Angles in the same segment are equal ( $∠$ s in the same seg). Angle in a semicircle is 90° ( $∠$ in semi-circle). Equal chords subtend equal angles (equal chords; equal $∠$ s). Cyclic quadrilaterals and tangents Opposite angles add to 180° (opp $∠$ s of cyclic quad). Exterior angle equals interior opposite angle (ext $∠$ of cyclic quad). Prove cyclic with a converse: opp $∠$ s supplementary, ext $∠$ = int opp $∠$ , or equal angles on the same side. Tangent is perpendicular to the radius (tan $⊥$ radius). Tangents from a common point are equal (tans from common pt). Tangent-chord angle equals the angle in the alternate segment (tan chord theorem). Proportion and similarity Line parallel to one side divides the other two in proportion (line $∥$ one side of $△$ ; name the lines). Midpoint theorem: joins midpoints, parallel to and half the third side (midpt theorem). Equal heights: area ratio = base ratio. Equiangular triangles are similar (reason: equiangular $△$ s); then sides are in proportion (reason: $∣∣∣$ $△$ s). Altitude in a right triangle: $A B^{2} = A D \cdot A C$ , $B D^{2} = A D \cdot D C$ . 'Hence prove' a product: cross-multiply the ratio from the similar triangles. Examinable proofs From Grade 11 (acute-angle case): line from centre to midpoint of chord is $⊥$ chord; angle at centre; opposite angles of a cyclic quad; tan-chord. From Grade 12: proportionality; equiangular triangles are similar; the perpendicular from the right angle to the hypotenuse gives similar triangles. Converses are not examinable as proofs. No concurrency and no proof by contradiction. Know by heart All reasons in their accepted short form The seven examinable proofs with their constructions On the formula sheet Nothing; all theorems and reasons must be known Common traps Statements without reasons lose marks. Using cyclic quad properties before proving the quad is cyclic. Naming similar triangles in the wrong order. Assuming a line is a diameter or tangent because it looks like one.
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IEB Grade 12 Euclidean Geometry Cheat Sheet

Euclidean geometry is about 40 marks of Paper 2. Every statement needs a reason, so this sheet gives the accepted short form for each.

Circle theorems (reason in brackets)

Line from centre perpendicular to a chord bisects it (line from centre $⊥$ chord).
Line from centre to midpoint of a chord is perpendicular to it (line from centre to midpt of chord).
Angle at the centre is twice the angle at the circumference ( $∠$ at centre $= 2 \times ∠$ at circumference).
Angles in the same segment are equal ( $∠$ s in the same seg).
Angle in a semicircle is 90° ( $∠$ in semi-circle).
Equal chords subtend equal angles (equal chords; equal $∠$ s).

Cyclic quadrilaterals and tangents

Opposite angles add to 180° (opp $∠$ s of cyclic quad).
Exterior angle equals interior opposite angle (ext $∠$ of cyclic quad).
Prove cyclic with a converse: opp $∠$ s supplementary, ext $∠$ = int opp $∠$ , or equal angles on the same side.
Tangent is perpendicular to the radius (tan $⊥$ radius).
Tangents from a common point are equal (tans from common pt).
Tangent-chord angle equals the angle in the alternate segment (tan chord theorem).

Proportion and similarity

Line parallel to one side divides the other two in proportion (line $∥$ one side of $△$ ; name the lines).
Midpoint theorem: joins midpoints, parallel to and half the third side (midpt theorem).
Equal heights: area ratio = base ratio.
Equiangular triangles are similar (reason: equiangular $△$ s); then sides are in proportion (reason: $∣∣∣$ $△$ s).
Altitude in a right triangle: $A B^{2} = A D \cdot A C$ , $B D^{2} = A D \cdot D C$ .
'Hence prove' a product: cross-multiply the ratio from the similar triangles.

Examinable proofs

From Grade 11 (acute-angle case): line from centre to midpoint of chord is $⊥$ chord; angle at centre; opposite angles of a cyclic quad; tan-chord.
From Grade 12: proportionality; equiangular triangles are similar; the perpendicular from the right angle to the hypotenuse gives similar triangles.
Converses are not examinable as proofs. No concurrency and no proof by contradiction.

Know by heart

All reasons in their accepted short form
The seven examinable proofs with their constructions

On the formula sheet

Nothing; all theorems and reasons must be known

Common traps

Statements without reasons lose marks.
Using cyclic quad properties before proving the quad is cyclic.
Naming similar triangles in the wrong order.
Assuming a line is a diameter or tangent because it looks like one.