In the previous section we saw some solved examples related to the centre of circles and chords. In this section, we will see 'equal chords'.
We know that the line joining
two points on a circle is called a chord. When we follow this rule, a
‘diameter’ is also a chord. It is the chord which passes through
the centre of the circle. In fact, it is the longest chord possible
in a given circle. This is shown in the fig.11(a) below.
 |
| Fig.17.11 Diameter is the longest chord |
All the other chords
which do not pass through the centre, are shorter than the diameter.
This is true for chords drawn in any direction, as shown in figs.17.11.a, b
and c.
• From the figs., we get a
feeling that chords which are situated at equal distances from the
centre are equal in length. We will prove it mathematically:
• Consider the circle in fig.17.12(a) below.
 |
| Fig.17.12 |
1. We have two yellow lines OP and OQ.
• They are of equal lengths
• They are drawn from the
centre.
• They are drawn in any two
convenient (nearly opposite) directions
2. A chord AB is drawn in such a
way that it is perpendicular to OP. This is shown in fig.b [ We can
draw a perpendicular to any given line using ruler and set square.
Here we draw a perpendicular to OP through ‘P’. Then we extend
the perpendicular both ways to meet the circle at A and B. Thus we
get the chord AB]
3. Similarly, a chord CD is drawn
in such a way that it is perpendicular to OQ. After completing fig.b,
we can say these:
• AB is at a perpendicular
distance of OP from the centre
• CD is at a perpendicular
distance of OQ from the centre.
Since OP and OQ are equal, We
can say:
■ The two chords AB and CD are
at the same perpendicular distance from the centre.
• We do not know whether AB and
CD are of the same length yet. But we will soon find out.
It is important to note
the significance of ‘perpendicular distance’. In the fig.c, OP’ and OP’’ are drawn. We could measure the distance from the
centre along those lines also. But those measurements will not be
‘perpendicular distances’. Perpendicular distance is the shortest distance. For our present discussion we consider
perpendicular lines OP and OQ only.
4. Now draw OA and OC. (fig.17.12.d)
5. Consider ΔOAP. It
is a right angled triangle. So applying Pythagoras theorem, we get:
AP = √[OA2 - OP2
]
6. The right side of the above
can be written as √[OC2 - OQ2 ] (∵ OA =
OC, and OP = OQ)
7. But when we consider the right
angled triangle OCQ, we find that √[OC2 - OQ2 ]
= CQ
8. So we can write: AP = √[OA2
- OP2 ] = √[OC2 - OQ2 ] = CQ. ⇒ AP = CQ
9. Now, what is AP?
• We have the chord AB, and OP
is the perpendicular drawn from the centre to that chord. So by
theorem 17.1, AP is half of AB
• Similarly, CQ is half of CD.
So from (8), we can write:
10. Half of AB = Half of CD (∵ AP = CQ)
11. So, 'full of AB' will be
equal to 'full of CD'
12. That is., AB = CD
• So we proved that Chords at
the same distance from the centre of the circle are equal in length.
The converse can also be written:
• Chords of equal length are at
equal perpendicular distances from the centre of the circle. Let us
prove this also:
1. In the fig.17.13(a) below, two equal chords
(in any two convenient directions) AB and CD are drawn.
 |
| Fig.17.13 |
• It is not
difficult to draw two such chords. First, from any convenient point
A, using any convenient radius, draw an arc, cutting the circle at B.
Join AB. This is our first chord.
• Now, from any other convenient
point ‘C’, draw an arc with the same radius AB, cutting the
circle at ‘D’. Join CD. This is our second chord. AB and CD will
be of the same length.
[Note that, in fig.17.12(a), we drew
OP and OQ first. In fig.17.13(a), we are drawing AB and CD first.]
2. Now, draw a perpendicular from
the centre ‘O’ to AB. We will call the foot of the perpendicular
as ‘P’.
[Drawing a perpendicular to
any given line can be done easily using ruler and set square. Here,
we draw a perpendicular to AB in such a way that, the perpendicular
passes through the centre ‘O’]
• Similarly draw the
perpendicular OQ to CD. This is shown in fig.17.13(b). We have to prove that OP = OQ
3. Now draw OA and OC. (fig.17.13.c)
5. Consider ΔOAP. It is a right angled triangle. So applying Pythagoras theorem, we get:
OP = √[OA2 - AP2 ]
6. The right side of the above can be written as √[OC2 - CQ2 ] (∵ OA = OC, and AP = CQ, according to theorem 17.1)
7. But when we consider the right angled triangle OCQ, we find that √[OC2 - CQ2 ] = OQ
8. So we can write: OP = √[OA2 - AP2 ] = √[OC2 - CQ2 ] = OQ. ⇒ OP = OQ
• So we proved that distances of 'equal chords' from the centre are the same.
We will write the above two findings as a theorem and it's converse
Theorem 17.3:
■ Chords which are at equal distances from the centre, are of equal length
Converse of the theorem:
■ Chords which are equal in length, are at equal distances from the centre
Based on the theorem 17.3
above, we can prove another property of chords.
We can write a theorem based on the above discussion:
Now we will show the proof for the converse:
We will now see a solved example
• In fig.17.14(a) below, AB and CD are
two equal chords. Then based on converse of theorem 17.3, the perpendicular
distances OP and OQ will be equal.
 |
| Fig.17.14 |
• If we extend the chords AB and
CD, they will meet at a point 'M' outside the circle. This is shown
in fig.17.14(b). The extension beyond the circle is shown in 'dashed line' for clarity.
• In fig.(c), a diameter is
drawn through 'O' in such a way that, when extended, it will pass
through 'M'. Now we can begin our calculations.
1. Consider the triangle OMP.
It is right angled. Applying Pythagoras theorem we get:
2. MP = √[MO2 -
OP2 ]. In this we can put OQ instead of OP. The result
will not change because OP = OQ. So we get:
3. MP = √[MO2 -
OQ2 ]
4. Now consider triangle OMQ.
It is also right angled. Applying Pythagoras theorem we get:
5. MQ = √[MO2 -
OQ2 ]. The right side is same as that in (3). So we can
write:
6. MP = √[MO2 -
OQ2 ] = MQ. That is., MP = MQ
8. Let us write the
correspondence:
9. M↔M, O↔O and P↔Q.
From M↔M, it is clear that the angles at M are equal. So we can
write ∠OMP = ∠OMQ
• So we can write: OM bisects
the angle at the point of intersection.
• In an earlier problem , we saw
that that the diameter bisects the angle between 'equal chords'. In
that problem, the two equal chords meet at a point on the circle.
• Now we find that the diameter
will bisect the angle between two equal chords, even if the chords
meet at a point outside the circle.
• In fact it is true even if the
two chords meet at a point in the interior of the circle. This is
shown in fig.17.15 below:
 |
| Fig.17.15 |
• In the fig.17.15, OP and OQ are equal. They are
perpendicular to AB and CD. So AB and CD are two equal chords
• The equal chords AB and CD
meet at 'M'
• The diameter (shown in red
line) bisects ∠CMB and ∠AMD
We can write a theorem based on the above discussion:
Theorem 17.4:
■ The angle at the meeting point
of two equal chords is bisected by the diameter through the meeting
point
The converse of this theorem
can also be written:
■ If the diameter through the
meeting point of two chords bisect the angle between those chords,
then the chords are of equal length.
Now we will show the proof for the converse:
• In the fig.17.16(a), any convenient
chord AB is drawn. In the fig.b, a convenient diameter (shown in red
colour) is drawn. Both the chord and the diameter are extended
towards the left to meet at M
 |
| Fig.17.16 |
• Measure ∠OMB. Draw a line MD (fig.c) in such a way that MD makes the same ∠OMB with the diameter. So we
have ∠DMO = ∠BMO
• Mark the point of intersection
of MD with the circle as 'C'. Make the portion between M and C 'dashed'. Now CD is a chord.
• We have two chords AB and CD.
We do not know whether they are equal in length. But we will soon
find out.
• We can say this: The diameter
bisects the angle between the two chords AB and CD (∵ ∠DMO =
∠BMO)
• Draw OP perpendicular to AB,
and OQ perpendicular to CD. We can now begin the calculations
1. Consider the triangle OMP.
It is right angled.
2. ∠MOP = [180 - (∠OMP +
90)] (∵ sum of interior angles in any triangle = 180o)
⇒∠MOP = (180 -∠OMP
– 90) = 90-∠OMP
3. Consider the triangle OMQ.
It is right angled.
4. ∠MOQ = [180 - (∠OMQ +
90)] (∵ sum of interior angles in any triangle = 180o)
⇒∠MOQ = (180 -∠OMQ
– 90) = 90-∠OMQ = 90-∠OMP (∵ ∠OMP = ∠OMQ)
• So from (2) and (4) we get
∠MOP = ∠MOQ
• Thus, in ⊿OMP, we have:
♦ one
side MO
♦ two angles at it's ends: ∠OMP and ∠MOP.
• This same combination is
present in ⊿OMQ also. It is a case of ASA congruence. Let us write
the correspondence: M↔M, O↔O and P↔Q
• Using O↔O and P↔Q we can
write: OP = OQ
• But OP and OQ are the
perpendicular distances from the centre. If they are equal, according
to theorem 17.3, the chords are also equal.So AB = CD
• Thus we have proved that, the chords are equal, and so this is the proof for the converse of theorem 17.4
• Thus we have proved that, the chords are equal, and so this is the proof for the converse of theorem 17.4
We will now see a solved example
Solved example 17.5
In the fig.17.17(a), the angle
between the radii and the chords are equal.
 |
| Fig.17.17 |
Prove that, the chords
are equal in length.
Solution:
In fig.17.7(b), the chords are
named. And also, two perpendicular lines OP and OQ are drawn. It is
given that ∠OAP = ∠OCQ. We can now begin the calculations
1. Consider the triangle OAP.
It is right angled.
2. ∠AOP = [180 - (∠OAP +
90)] (∵ sum of interior angles in any triangle = 180o)
same as ∠AOP = (180 -∠OAP
– 90) = 90-∠OAP
3. Consider the triangle OCQ.
It is right angled.
4. ∠COQ = [180 - (∠OCQ +
90)] (∵ sum of interior angles in any triangle = 180o)
⇒ ∠COQ = (180 -∠OCQ
– 90) = 90-∠OCQ = 90-∠OAP (∵ ∠OAP = ∠OCQ)
5. So from (2) and (4) we get
∠AOP = ∠COQ
6. Thus, in ⊿OAP, we have:
♦ one side AO and
♦ two angles at it's
ends: ∠OAP and ∠AOP.
7. This same combination is
present in ⊿OCQ also:
♦ one side CO and
♦ two angles at it's
ends: ∠OCQ and ∠COQ.
8. But the two combinations
are the same as shown below:
♦ sides AO = CO (since
radii of the same circle)
♦ ∠OAP = ∠OCQ
(given)
♦ ∠AOP = ∠COQ (from
(5))
9. It is a case of ASA
congruence. Let us write the correspondence:
O↔O, A↔C and P↔Q
10. Using A↔C and P↔Q we can
write: AP = CQ
11. But AP and CQ are ‘half
chords’. Because, according to theorem 17.1, the perpendicular from
the centre bisects the chord
12. So we can write: Half of AB =
Half of CD
Then obviously, Full of AB =
Full of CD. That is., AB = CD
Thus we have proved that, the
chords are equal.
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