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Lesson 5 Theorem

The tangents to a circle from an external point are equal in length. If a straight line touches a circle, and from the point of contact a chord is drawn, the angles which the chord makes with the tangent are equal to the angles in the alternate segments (on the other side of the chord from the tangent). In two examples, angles are calculated using properties of tangents, alternate segments, and isosceles triangles.

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Monique Newen
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344 views5 pages

Lesson 5 Theorem

The tangents to a circle from an external point are equal in length. If a straight line touches a circle, and from the point of contact a chord is drawn, the angles which the chord makes with the tangent are equal to the angles in the alternate segments (on the other side of the chord from the tangent). In two examples, angles are calculated using properties of tangents, alternate segments, and isosceles triangles.

Uploaded by

Monique Newen
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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Tangents from an external point

Theorem

The tangents to a circle from an external point are equal

___ ___
Given: a point T outside a circle, centre O, TA and TB are tangents to the circle at A and B.
A

Fig. 29 B

To prove: |TA| = |TB|.

___ ____ ____


Construction: Join OA, OB and OT

Proof:

In ∆ s OAT and OBT,

<A = <B = 90° (radius tangent)

|OA| = |OT| (common side)

∆ OAT = ∆ OBT (R.H.S.)

|TA| = |TB|

Notice also that <AOT = <BOT and <ATO = <BTO.


Hence the line joining the external point to the centre of the circle bisects the angle between the

tangents and the angle between the radii drawn to the points of contact of the tangents, i.e. OT is

on the line of symmetry of fig. 29.

ALTERNATE SEGMENT

____
In both parts of fig. 30 SAT is a tangent to the circle at A. The chord AB divides the circle into

two segments APB and AQB.

P.
a) B

. Q

S A T

b)
P.
B

.Q

S A T
Fig. 30

___
In fig. 30 (a) the segment APB is the alternate segment to <TAB, i.e. it is on the other side of

AB from <TAB. Similarly, in fig. 30(b), segment AQB is the alternate segment to <SAB.

Sometimes the word opposite is used instead of alternate.

Alternate segment theorem

Theorem
If a straight line touches a circle, and from the point of contact a chord is drawn< the angles

which the chord makes with the tangent are equal to the angles in the alternate segments.

____
Given: a circle, with SAT a tangent at A and chord AB dividing the circle into two segments

APB and AQB. Segment APB is alternate to <TAB.


D

P x3 x2

y x1 Q

S A T

Fig. 31

To prove: <TAB = <APB and <SAB = <AQB.

Construction: draw the diameter AD. Join BD.

Proof:

With the lettering of fig. 31,

x1 + y = 90° (tangent radius)

also <ABD = 90° (angle in semicircle)

∴x2 + y = 90° (sum of angles of ∆ )

∴x1= x2 =x3 (angles in the same segment)

∴ <TAB = <APB

Also <SAB = 180° - x1 (angles on a straight line)


= 180° - x3 (x1 = x3 proved)

= <AQB (opposite angles of cyclic quadrilateral)

Example 1

In fig. 32 O is the centre of the circle and TA and TB are tangents.

a) If <ATO = 36°, calculate <ABO.

b) If <ABT = 57°, calculate <AQT.

T
X O

fig.32 B

Solution
a) <BTO = 36°
∴<ABO = 90° - 36°
b)If <ATO = 57° then <OBA = 33°
∴ <AOT = 90° - 33°= 57°
Example 2
In the fig. 33, TY is tangent to the circle TVS. If <SVT = 48° and |VS| = |ST|, what is
<VTY?

S
48° V

fig. 33 T Y
Solution
<VTS = 48° (base angles of isosceles ∆ )
<TSV = 180° - (2× 48°) = 84°
∴ <VTY = 84° (alternate to <TSV)

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