Circles iv
Find the values of the unknown letters.
Theorem 9 Two tangents drawn to a circle from the same point outside the circle are equal in length.
Proof :
Consider a circle, with centre
O . Choose a point
P outside the circle. Draw two tangents to the circle from point
P , that meet the circle at
A and
B . Draw lines
O
A ,
O
B and
O
P .
The aim is to prove that
A
P
=
B
P . In
▵
O
A
P and
▵
O
B
P ,
O
A
=
O
B (radii)
∠
O
A
P
=
∠
O
P
B
=
90
∘ (
O
A
⊥
A
P and
O
B
⊥
B
P )
O
P is common to both triangles.
▵
O
A
P
≡
▵
O
B
P (right angle, hypotenuse, side)
∴
A
P
=
B
P
Circles v
Find the value of the unknown lengths.
Theorem 10 The angle between a tangent and a chord, drawn at the point of contact of the chord, is equal to the angle which the chord subtends in the alternate segment.
Proof :
Consider a circle, with centre
O . Draw a chord
A
B and a tangent
S
R to the circle at point
B . Chord
A
B subtends angles at points
P and
Q on the minor and major arcs, respectively.
Draw a diameter
B
T and join
A to
T .
The aim is to prove that
A
P
B
^
=
A
B
R
^ and
A
Q
B
^
=
A
B
S
^ . First prove that
A
Q
B
^
=
A
B
S
^ as this result is needed to prove that
A
P
B
^
=
A
B
R
^ .
A
B
S
^
+
A
B
T
^
=
90
∘
(
TB
⊥
SR
)
B
A
T
^
=
90
∘
(
∠
's at centre
)
∴
A
B
T
^
+
A
T
B
^
=
90
∘
(
sum of angles in
▵
BAT
)
∴
A
B
S
^
=
A
B
T
^
However,
AQB
^
=
A
T
B
^
(
angles subtended by same chord
AB
)
∴
A
Q
B
^
=
A
B
S
^
S
B
Q
^
+
Q
B
R
^
=
180
∘
(
SBT
is a str. line
)
A
P
B
^
+
A
Q
B
^
=
180
∘
(
ABPQ
is a cyclic quad
)
∴
S
B
Q
^
+
Q
B
R
^
=
A
P
B
^
+
A
Q
B
^
AQB
^
=
A
B
S
^
∴
A
P
B
^
=
A
B
R
^
Circles vi
Find the values of the unknown letters.
Theorem 11 (Converse of
[link] ) If the angle formed between a line, that is drawn through the end point of a chord, and the chord, is equal to the angle subtended by the chord in the alternate segment, then the line is a tangent to the circle.
Proof :
Consider a circle, with centre
O and chord
A
B . Let line
S
R pass through point
B . Chord
A
B subtends an angle at point
Q such that
A
B
S
^
=
A
Q
B
^ .
The aim is to prove that
S
B
R is a tangent to the circle. By contradiction. Assume that
S
B
R is not a tangent to the circle and draw
X
B
Y such that
X
B
Y is a tangent to the circle.
A
B
X
^
=
A
Q
B
^
(
tan
-
chord theorem
)
However
,
ABS
^
=
A
Q
B
^
(
given
)
∴
A
B
X
^
=
A
B
S
^
But
,
ABX
^
=
A
B
S
^
+
X
B
S
^
can only be true if
,
XBS
^
=
0
If
X
B
S
^ is zero, then both
X
B
Y and
S
B
R coincide and
S
B
R is a tangent to the circle.
Applying theorem
[link]
Show that Theorem
[link] also applies to the following two cases:
B
D is a tangent to the circle with centre
O .
B
O
⊥
A
D .
Prove that:
C
F
O
E is a cyclic quadrilateral
F
B
=
B
C
▵
C
O
E
/
/
/
▵
C
B
F
C
D
2
=
E
D
.
A
D
O
E
B
C
=
C
D
C
O
To show a quadrilateral is cyclic, we need a pair of opposite angles to be supplementary, so let's look for that.
F
O
E
^
=
90
∘
(
BO
⊥
OD
)
F
C
E
^
=
90
∘
(
∠
subtended by diameter
AE
)
∴
C
F
O
E
is a cyclic quad
(
opposite
∠
's supplementary
)
Since these two sides are part of a triangle, we are proving that triangle to be isosceles. The easiest way is to show the angles opposite to those sides to be equal.
Let
O
E
C
^
=
x .
∴
F
C
B
^
=
x
(
∠
between tangent
BD
and chord
CE
)
∴
B
F
C
^
=
x
(
exterior
∠
to cyclic quad
CFOE
)
∴
B
F
=
B
C
(
sides opposite equal
∠
's in isosceles
▵
BFC
)
To show these two triangles similar, we will need 3 equal angles. We already have 3 of the 6 needed angles from the previous question. We need only find the missing 3 angles.
C
B
F
^
=
180
∘
-
2
x
(
sum of
∠
's in
▵
BFC
)
O
C
=
O
E
(
radii of circle
O
)
∴
E
C
O
^
=
x
(
isosceles
▵
COE
)
∴
C
O
E
^
=
180
∘
-
2
x
(
sum of
∠
's in
▵
COE
)
C
O
E
^
=
C
B
F
^
E
C
O
^
=
F
C
B
^
O
E
C
^
=
C
F
B
^
∴
▵
C
O
E
|||
▵
C
B
F
(
3
∠
's equal
)
This relation reminds us of a proportionality relation between similar triangles. So investigate which triangles contain these sides and prove them similar. In this case 3 equal angles works well. Start with one triangle.
In
▵
E
D
C
C
E
D
^
=
180
∘
-
x
(
∠
's on a str. line
AD
)
E
C
D
^
=
90
∘
-
x
(
complementary
∠
's
)
Now look at the angles in the other triangle.
In
▵
A
D
C
A
C
E
^
=
180
∘
-
x
(
sum of
∠
's
ACE
^
and
ECO
^
)
C
A
D
^
=
90
∘
-
x
(
sum of
∠
's in
▵
CAE
)
The third equal angle is an angle both triangles have in common.
Lastly,
A
D
C
^
=
E
D
C
^ since they are the same
∠ .
Step 4d: Now we know that the triangles are similar and can use the proportionality relation accordingly.
∴
▵
A
D
C
|||
▵
C
D
E
(
3
∠
's equal
)
∴
E
D
C
D
=
C
D
A
D
∴
C
D
2
=
E
D
.
A
D
This looks like another proportionality relation with a little twist, since not all sides are contained in 2 triangles. There is a quick observation we can make about the odd side out,
O
E .
O
E
=
C
D
(
▵
OEC
is isosceles
)
With this observation we can limit ourselves to proving triangles
B
O
C and
O
D
C similar. Start in one of the triangles.
In
▵
B
C
O
O
C
B
^
=
90
∘
(
radius
OC
on tangent
BD
)
C
B
O
^
=
180
∘
-
2
x
(
sum of
∠
's in
▵
BFC
)
Then we move on to the other one.
In
▵
O
C
D
O
C
D
^
=
90
∘
(
radius
OC
on tangent
BD
)
C
O
D
^
=
180
∘
-
2
x
(
sum of
∠
's in
▵
OCE
)
Again we have a common element.
Lastly,
O
C is a common side to both
▵ 's.
Step 5e: Then, once we've shown similarity, we use the proportionality relation , as well as our first observation, appropriately.
∴
▵
B
O
C
|||
▵
O
D
C
(
common side and 2 equal angles
)
∴
C
O
B
C
=
C
D
C
O
∴
O
E
B
C
=
C
D
C
O
(
OE
=
CD
isosceles
▵
OEC
)
F
D is drawn parallel to the tangent
C
B Prove that:
F
A
D
E is cyclic
▵
A
F
E
|||
▵
C
B
D
F
C
.
A
G
G
H
=
D
C
.
F
E
B
D
In this case, the best way to show
F
A
D
E is a cyclic quadrilateral is to look for equal angles, subtended by the same chord.
Let
∠
B
C
D
=
x
∴
∠
C
A
H
=
x
(
∠
between tangent BC and chord CE
)
∴
∠
F
D
C
=
x
(
alternate
∠
,
FD
∥
CB
)
∴
FADE
is a cyclic quad
(
chord FE subtends equal
∠
's
)
To show these 2 triangles similar we will need 3 equal angles. We can use the result from the previous question.
Let
∠
F
E
A
=
y
∴
∠
F
D
A
=
y
(
∠
's subtended by same chord
AF
in cyclic quad
FADE
)
∴
∠
C
B
D
=
y
(
corresponding
∠
's,
FD
∥
CB
)
∴
∠
F
E
A
=
∠
C
B
D
We have already proved 1 pair of angles equal in the previous question.
∠
B
C
D
=
∠
F
A
E
(
above
)
Proving the last set of angles equal is simply a matter of adding up the angles in the triangles. Then we have proved similarity.
∠
A
F
E
=
180
∘
-
x
-
y
(
∠
's in
▵
AFE
)
∠
C
B
D
=
180
∘
-
x
-
y
(
∠
's in
▵
CBD
)
∴
▵
A
F
E
|||
▵
C
B
D
(
3
∠
's equal
)
This equation looks like it has to do with proportionality relation of similar triangles. We already showed triangles
A
F
E and
C
B
D similar in the previous question. So lets start there.
D
C
B
D
=
F
A
F
E
∴
D
C
.
F
E
B
D
=
F
A
Now we need to look for a hint about side
F
A . Looking at triangle
C
A
H we see that there is a line
F
G intersecting it parallel to base
C
H . This gives us another proportionality relation.
A
G
G
H
=
F
A
F
C
(
FG
∥
CH
splits up lines
AH
and
AC
proportionally
)
∴
F
A
=
F
C
.
A
G
G
H
We have 2 expressions for the side
F
A .
∴
F
C
.
A
G
G
H
=
D
C
.
F
E
B
D
