Circle Geometry (Part 3)

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Exam Paper 3

Circle Geometry (Part 3)

Assessment Standard: 12.4.1(c)

Cyclic Quadrilaterals

Last week we covered up toTheorem 3, the idea of a converse and we applied our theory to some problems called RIDERS. Okay, so now onto the next chunk of theory which has to do with CYCLIC

QUADRILATERALS.

Now remember - a Quadrilateral is any

four sided figure.

A Cyclic Quadrilateral is a four sided

figure with all four corners lying on the circumference of a circle. We usually name it clockwise or anti-clockwise -don’t ZIGZAG.

Theorem 5:

The opposite angles of a cyclic quadrilateral are supplementary (opposite ∠s of a cyclic quad) and conversely, if a pair of opposite angles of a quadrilateral is supplementary then the quad is cyclic.

(Opposite ∠s of quad suppl.)

D Not cyclic PSTQ P S Q T A Cyclic quad AbCD/DAbC/CbAD C b Not cyclic WQPR W _Q R _P Cyclic quad PQRS P Q S _R

Collect your Paper 3 Lessons every week!!

Guys, both NSC and IEb examinations

candidates have the option of writing Paper 3 at the end of the year! Paper 3 covers additional

mathematics material and is out of 100 marks. Maths Paper 3 will really set you apart in the job market, and make studying technical subjects at tertiary level easier. We have hooked you up with these lessons - written by IEb Maths

Paper 3 examiner, Heather Frankiskos.

The Examination Guidelines for the DOE stipulate that the proofs of all Grade 11 and the Grade 12 theorems of the geometry of circles are required for examination purposes.

This bookwork has a suggested maximum mark allocation of 15 out of 100. The IEb Subject Assessment Guidelines suggest that no bookwork i.e. NO PROOFS be examined.

We can expect ‘Complete the statement’ type questions but no complete proofs. These Geometry modules will not

present proofs. You must get these from your teacher.

Exam Paper 3

Circle Geometry (Part 3)

If AbCD cyclic then ^A + ^C = 180º ^b + ^D = 180º

If ^A + ^C = 180º then AbCD cyclic ^b + ^D = 180º

Theorem 6:

The exterior angle of a cyclic quad is equal to the interior opposite angle (ext ∠of cyclic quad) and conversely, if the exterior angle of a quad is equal to the interior opposite angle, the quad is cyclic (ext ∠ of quad)

If AbCD cyclic then ^C ₁= ^A

If ^C ₁ = ^A then AbCD is cyclic (we can draw a circle through all 4 points)

A b D C A b D C x 180 - x A b D C 1 x x A b x 80º

Exam Paper 3

Circle Geometry (Part 3)

Can you see the following?

1) ^E ₂= ^b ₁ (∠s same segment)

^

F ₁ = ^C (∠s same segment)

2) ^F ₁ + ^P ₁ = 180º (opp ∠s cyclic quad PEFb)

^

C + ^P ₁ = 180º (opp ∠s cyclic quad PECb)

^

E ₁+ ^b ₁+ ^b ₂ = 180º (opp ∠s cyclic quad ECbP)

^

E ₁+ ^E ₂+ ^b ₂ = 180º (opp ∠s cyclic quad EFbP) 3) ^P ₂ = ^F ₁ (ext ∠ cyclic quad PEFb)

^

b ₃ = ^E ₁ (ext ∠cyclic quad PbCE)

^

F ₂ = ^ P ₁ (ext ∠ cyclic quad FEPb)

PLEASE look at this RIDER carefully - use your fingers or colour (or whatever) to try to see all the angles mentioned above.

Also remember that you have met an exterior angle before. Do you remember that for any triangle, the exterior angle (formed by extending any side) was equal to the sum of the interior opposite two angles?

So ^b ₁ = ^A + ^C

Here for a cyclic quadrilateral the exterior angle is equal to the one interior opposite angle.

^

b ₁ = ^D Note: ^A ₁≠ ^C

bAF must be straight like AbK

C A b 1 C A b 1 A b C 1 C A b D x 1 1 K F K G E F P b 1 C 2 2 3 1 1 2 2 1

Exam Paper 3

Circle Geometry (Part 3)

The next bit of theory relates to a very special line called a TANGENT

We define a TANGENT to be a line which makes contact with a circle at only one point on the circumference. In each Rider below SPT is a tangent.

In each of the above sketches, draw in the radius OP. Can you see that OP will always be perpendicular to SPT?

This is your next theorem. Theorem 7:

A tangent to a circle is always perpendicular to the radius at the point of contact (rad ⊥ tangent) and conversely, a line drawn perpendicular to a radius at the point where it meets the circumference, will be a tangent (radius ⊥ to line)

Given SPT a tangent with OP a radius then OP ⊥ SPT

and conversely,

Given radius OP and SPT ⊥ OP then SPT is a tangent

P T S P T S _S P _T P T S

Exam Paper 3

Circle Geometry (Part 3)

Theorem 8:

If two tangents are drawn to a circle from a common external point, then the two distances from this point to the points of contact are equal (tang, same point)

And now for the final, but the most used theorem in Circle Geometry. We call it the TAN-CHORD theorem This theorem says

Theorem 9:

The angle formed between a tangent to a circle and a chord drawn from the point of contact.... chord drawn at point of contact angle between tangent and chord

is equal to the angle subtended by the same chord in the other segment (tan chord theorem)

x is the angle between tangent and chord PM coming in from the right

y is the angle between tangent and chord Pb coming in from the left

P Q R PR = PQ P Q R

PR ≠ PQ (both must be tangents from P)

PQ is actually called a secant

tangent to circle y M b x x y P T C

Exam Paper 3

Circle Geometry (Part 3)

In the diagram QTS and SPW are tangents Can you agree to all these statements? 1) ^T ₁ = ^P ₁ (tan chord bT) ^T ₄ = ^P ₃ (tan chord TK) ^P ₂ = ^T ₂ (tan chord bP) ^P ₄ = ^T ₃ (tan chord PK) 2) ^T ₁ + ^T ₂ = ^K (tan chord TP) ^T ₃ + ^T ₄ = ^b (tan chord TP) ^P ₁ + ^P ₂ = ^K (tan chord TP) ^P ₃ + ^P ₄ = ^b (tan chord TP) 3) TS = SP (tangents same PT) \ ^T ₃ + ^T ₄ = ^P ₃ + ^P ₄ (Isos ∆)

4) ^b + ^K = 180° (opp ∠s cyclic quad)

^

P ₁ + ^P ₃ + ^T ₂ + ^T ₃ = 180° (opp ∠s cyclic quad) 5) There are no exterior angles of cyclic quad bTKP in the diagram

6) There is no radius / diameter/ centre so no angle is 90° Of course this very important Theorem must have a converse...

If the angle formed between a chord and a line is equal to an angle subtended by the chord, then the line is a tangent to the circle (converse tan chord)

A C x 1x b T b W 3 S Q 4 1 2 P K 1 3 4 2 T

Exam Paper 3

Circle Geometry (Part 3)

Remember to read the given information slowly; pay attention to keywords such as centre, radius, cyclic quad, tangent and relate them to their theorems. Mark all information on picture.

We will start with an easy one Rider 1

Given:

P, Q, R and S are points on a circle. PR is the diameter of the circle MRT is a tangent at R

^S ₂ = 65° and ^R ₃ = 35°

Determine, with reasons, the sizes of ^R ₁; ^R ₄ and ^S ₃

Solution a) ^R ₁ = ^S ₂ (tan chord QR) \ ^R ₁ = 65° b) ^R ₃ + ^R ₄ = 90° (rad ⊥ tang) \ ^R ₄ = 55° c) ^S ₁ + ^S ₂ = 90° (∠ in semi-circle) \ ^S ₃ = 90° (∠on st line)

You were not asked to find any other angles but see if you get

^

P ₁ = 65°; ^S ₁ = 25° ; R ^₂ = 25°; ^T = 35°; ^Q ₁ = 35° etc.

Think: PQRS must be cyclic / PR is a diameter so

^

S ₁ + ^S ₂ = 90° /PR must be ⊥ to RT the tangent

Q M 3 35° 1 1 2 2 1 12 3 4 R S P 2 T 65°