Euclidean Geometry

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A B P O 2x x

Euclidean geometry in Grade 12 is the geometry of the circle. You will learn a set of theorems, and then use them to prove results (“riders”).

In this lesson you will learn to:

• name every part of a circle;
• use the chord theorems;
• apply the angle-at-the-centre theorem;
• use the angle in a semicircle and angles in the same segment;
• work with cyclic quadrilaterals;
• apply the tangent theorems;
• use the tangent-chord (alternate segment) theorem; and
• write a clear geometric proof with reasons.

1. The Circle

A B O M

A B C D

• A chord divides the circle into two segments — a major and a minor segment.
• Points that all lie on the same circle are concyclic.
• A cyclic quadrilateral has all four vertices on the circle.

\[ \text{name the parts precisely} \]

Key idea: every theorem that follows is stated in this vocabulary. Get the words exact and the reasons become easy to quote.

2. Chords

A B O M

Theorem: the line from the centre perpendicular to a chord bisects the chord.

A B O M

Converse: the line from the centre to the midpoint of a chord is perpendicular to the chord.

A useful consequence: the perpendicular bisector of any chord passes through the centre — that is how the centre of a circle is found from two chords.

\[ OM \perp AB \;\Longleftrightarrow\; AM = MB \]

Key idea: perpendicular-from-centre and bisects-the-chord always come together. Given one, you may state the other with the reason.

3. Angle at the Centre

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Theorem: the angle subtended by an arc at the centre is twice the angle subtended by the same arc at the circumference.

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If \(A\hat{P}B = 58^\circ\), find the reflex and non-reflex angle at the centre.

Always check whether the question wants the angle on the same side as \(P\) or the reflex angle on the other side.

\[ \text{centre angle} = 2 \times \text{circumference angle} \]

Key idea: both angles must stand on the SAME arc. This one theorem is the parent of the next three.

4. Angles in a Circle

A B P 90°

Theorem: the angle in a semicircle is a right angle.

A B P Q

Theorem: angles in the same segment, standing on the same arc, are equal.

\[ \text{semicircle} \Rightarrow 90^\circ \qquad \text{same segment} \Rightarrow \text{equal} \]

Key idea: both of these are children of the angle-at-the-centre theorem. If you forget them, you can rederive them from it.

5. Cyclic Quadrilaterals

A B C D

Theorem: the opposite angles of a cyclic quadrilateral are supplementary (add to \(180^\circ\)).

A B C D

Theorem: the exterior angle of a cyclic quadrilateral equals the interior opposite angle.

This follows directly from the opposite-angles theorem, since the exterior angle and its own interior angle also add to \(180^\circ\).

A B C D

To prove four points lie on a circle, prove ANY one of:

• a pair of opposite angles is supplementary;
• an exterior angle equals the interior opposite angle;
• two points subtend equal angles on the same side of a line (same-segment converse).

\[ \hat{A} + \hat{C} = 180^\circ \]

Key idea: opposite angles supplementary, and exterior = interior opposite. The converses prove concyclic points.

6. Tangents

T O

Theorem: a tangent is perpendicular to the radius at the point of contact.

A B O P

Theorem: two tangents drawn from the same external point are equal in length.

It also follows that the line from the external point to the centre bisects the angle between the tangents, and bisects the angle \(A\hat{O}B\) at the centre.

\[ \text{tan} \perp \text{radius} \qquad PA = PB \]

Key idea: the right angle at the point of contact, and the two equal tangents, unlock most tangent problems. Look for the radius first.

7. Tangent-Chord

A B C

Theorem: the angle between a tangent and a chord equals the angle in the alternate segment.

A B C

This theorem is the one learners miss most. Look for it whenever you see:

• a tangent line, AND
• a chord drawn from the point of contact.

\[ \text{tangent-chord angle} = \text{angle in alternate segment} \]

Key idea: tangent + chord from the contact point = tan-chord. Always look across the circle to the alternate segment.

8. Proving Riders

\[ \text{statement} \;+\; \text{reason} \]

A geometry proof is a chain, and every link needs a reason.

1. Mark everything you know on the diagram (equal sides, right angles, given angles).
2. Decide what you must prove, and work backwards from it.
3. Write each step as a STATEMENT with its REASON in brackets.
4. Quote reasons exactly — “tan ⊥ radius”, “∠s same segment”, etc.

A B P 90°

\(AB\) is a diameter and \(P\) is on the circle. Prove \(A\hat{P}B = 90^\circ\).

Situation Reason Line from centre bisects chord line from centre ⊥ chord Angle at centre ∠ at centre = 2 ∠ at circumference Diameter subtends the angle ∠ in semi-circle Two angles on the same chord ∠s in same segment Opposite angles of cyclic quad opp ∠s of cyclic quad Tangent and radius tan ⊥ radius Two tangents from a point tangents from common point Tangent and chord tan-chord

\[ \text{every statement earns a reason} \]

Key idea: in the examination, the reason column is where most marks are won or lost. Never write a statement without its reason.

A B P O 2x x

The circle theorems, in one place:

• line from centre ⊥ chord bisects it (and its converse);
• angle at centre \(= 2 \times\) angle at circumference;
• angle in a semicircle \(= 90^\circ\);
• angles in the same segment are equal;
• opposite angles of a cyclic quad are supplementary; exterior = interior opposite;
• tangent \(\perp\) radius; two tangents from a point are equal;
• tangent-chord angle = angle in the alternate segment.