Chapter 5: Circles - Theorems and Proofs

Flashcards covering the theorems, proofs, and congruence criteria for circle properties in Chapter 5: Circles.

Theorem 1: Equal Chords Subtend Equal Angles

States that equal chords of a circle subtend equal angles at the centre. For a circle with centre $O$, if chord $AB = CD$, then $\times$ is not needed here; $\triangle AOB$ and $\triangle OCD$ give $\text{CPCT}$ result $\text{m}\times \text{or } \text{angle } \text{measurement } \rightarrow \text{angle} \text{ equality } \rightarrow \text{i.e., } \text{chord equality yields } \text{angle equality: } \text{chord } AB = \text{chord } CD \rightarrow \text{angle } AOB = \text{angle } COD$.

Proof of Theorem 1

In $\triangle OAB$ and $\triangle OCD$: $AB = CD$ (given), $OB = OC$ (radii of same circle), and $OA = OD$ (radii of same circle). By $SSS$ criteria, $\triangle OAB \triangleq \triangle OCD$ (or $\triangle OAB \text{ is congruent to } \triangle OCD$), which proves $\text{angle } AOB = \text{angle } COD$ by cpct.

Theorem 2: Equal Angles Subtend Equal Chords

States that if the angles subtended by the chords at the centre are equal, then the chords are equal. For a circle with centre $O$, if $\text{angle } AOB = \text{angle } COD$, then $AB = CD$.

Proof of Theorem 2

In $\triangle OAB$ and $\triangle OCD$: $OA = OD$ (radii of same circle), $\text{angle } AOB = \text{angle } COD$ (given), and $OB = OC$ (radii of same circle). By $SAS$ criteria, $\triangle AOB \text{ is congruent to } \triangle OCD$, which proves $AB = CD$ by cpct.

Theorem 3: Perpendicular from Centre to a Chord

States that the perpendicular from the centre of a circle to a chord bisects the chord. For a circle with centre $O$ and chord $AB$, if $OD \text{ is perpendicular to } AB$ ($\text{angle } ODA = \text{angle } ODB = 90^\text{o}$), then $AD = BD$.

Proof of Theorem 3

In $\triangle ODA$ and $\triangle ODB$: $\text{angle } ODA = \text{angle } ODB = 90^\text{o}$ (each $90^\text{o}$), $OA = OB$ (radii of same circle), and $OD = OD$ (common). By $RHS$ criteria, $\triangle ODA \text{ is congruent to } \triangle ODB$, which proves $AD = BD$ by cpct.

CPCT

Stands for 'corresponding parts of congruent triangles', used to conclude that corresponding angles or sides of two proven congruent triangles are equal.