In the previous section we saw the method to determine the centre of any circle or portions of any circle. In this section, we will discuss more about Circles.
Consider the circle and the
triangle inside it, in the fig.17.4(a) below (right click and select 'open in new tab' for an enlarged view). Note the following two points
carefully:
• The triangle is an equilateral triangle. That is., all it’s sides are equal
• The circle passes through all
the 3 vertices of the triangle.
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| Fig.17.4 |
Now, we don’t know yet, how
to draw such a figure. That is.,
♦ If we are given an equilateral
triangle, we do not know how to draw a circle, passing through all
it’s 3 vertices. OR
♦ If we are given a circle, we
do not know how to draw an equilateral triangle inside it.
• Someone gave us the
fig.17.4(a). We are now going to analyse it, and find a method to draw such a figure by ourselves.
• Consider fig.b. The given
triangle has been named as ABC. Also, 3 inner triangles are drawn, by
joining the centre to the 3 vertices. We get 3 inner triangles: OAB,
OBC, and OCA
• Now we will prove that, these
three triangles are equal.
• First consider ΔOAB and ΔOBC
♦ OB = OB ♦ OA = OC (∵ radii of the same circle) ♦ AB = BC (∵ sides of equilateral ΔABC)
• In a similar way we can prove
ΔOAB is congruent to ΔOAC also. So we can write:
• ΔOAB = ΔOBC AND ΔOAB =ΔOAC.
From this we get:
■ ΔOAB = ΔOBC = ΔOAC
• The angle at O is common for all the 3 equal triangles. So the 3 angles at O are equal.
• The total angle at O is 360o .
• If each angle is denoted as 'x', we can write: 3x = 360o. From this we get x = 120o.
■ Thus we get each of the 3 angles
at the centre O as 120o degrees
• Now we want to draw a radius
of the circle. It must satisfy one condition.
■ The condition is: The
radius must be perpendicular to any one of the sides of the
equilateral triangle ABC.
• In the fig.c, the radius OP is
drawn which satisfies the above condition.
[Drawing such a radius is not
at all difficult. We know how to draw a perpendicular (through any
given point) to any given line using set square. Here, we want to
draw a perpendicular to AB through ‘O’. Once it is drawn, extend
it to meet the circle at P]
• We drew OP perpendicular to the base AB. So OP
will bisect ∠AOB
• Thus we get ∠AOP = ∠BOP =
60o.
• Now join A and P to form ΔOAP.
Here we have an interesting situation:
• We have ΔOAP, and we have the
following two information about it:
♦ It’s two sides OA and OP
are equal (∵ they are radii of the same circle)
♦ The ∠AOP between those
equal sides is 60o .
• If in a triangle, two sides
are equal, and the angle between them is 60o, we can
obtain the complete details about that triangle. The steps are shown
below:
• One angle is 60o.
So the sum of the other two angles OAP and OPA is 180 – 60 = 120o.
♦ Spread of ∠OAP is OP
♦ Spread of ∠OPA is OA
• But OP = OA. That is., spreads
are equal. So angles are equal. That is., ∠OAP = ∠OPA
So we have two equal angles
and their sum is 120o.
That is., x + x = 120o. ⇒ 2x = 120o ⇒ x = 60o.
So all the three angles of
ΔOAP are equal to 60o. Thus, it is an equilateral
triangle.
This (see fig.17.4.d) is an important
result to remember. We will write it in the form of a theorem:
Theorem 17.2:
Theorem 17.2:
■ If it is given that two sides
of a triangle are equal, and the angle between them is 60o,
then it is an equilateral triangle.
• Now consider the
equilateral ΔOAP. Consider it as having the base OP and apex
A.
• We have a line AB passing
through the apex A, and at the same time perpendicular to base OP.
■ In any equilateral triangle,
if a line passing through the apex is perpendicular to the base, then
that perpendicular line will bisect the base.
• So OP is bisected by AB. And
this leads us to the following result:
■ Any side (in our case, side
AB) of an equilateral triangle drawn inside a circle, will bisect the
radius (in our case, radius OP) drawn perpendicular to that side.
Is the reverse true? That is:
3. We draw a perpendicular
bisector XY of that radius OP. The radius OP is bisected at 'M' (fig.7.5.a)
4. We extend the perpendicular
bisector both ways to meet the circle at A and B (fig.7.5.b)
5. When the perpendicular
bisector meets the circle at it’s ends, it become a chord of the
circle
■ Is this chord, a side of the equilateral triangle that we are seeking?
Let us find out:
1. If this chord AB is indeed a
side of the equilateral triangle, we have obtained two vertices A and
B. 2. We need to find only the third vertex C
3. This third vertex will lie on
the perpendicular bisector of the side AB.
4. In the fig.b, it is clear that
OP is perpendicular to AB. But is it a bisector also?
5. That is., we want OP to
satisfy two conditions:
• It must be perpendicular to AB
• It must bisect AB
It indeed will bisect AB according to theorem 17.1:
■ A perpendicular from the
centre to a chord will bisect the chord. So OP is indeed the
perpendicular bisector of AB
6. So the third vertex C will lie
some where on OP. But where?
7. We want all the vertices of
the equilateral triangle ABC to lie on the circle.
8. So extend OP to meet the
circle. That meeting point will be our required point C
9. Join A to C and also B to C. Triangle ABC is complete.
Using a divider, we can check that the three sides AB, BC and CA are
equal.
But a check using divider is
not enough. We must prove mathematically that ΔABC is an equilateral
triangle
It can be done as follows:
We want to prove that ΔABC in the above fig.17.5(c) is an equilateral triangle. This fig. is reproduced in 17.6(a) below (right click and select 'open in new tab' for an enlarged view).
Let us begin the calculation steps:
We want to prove that ΔABC in the above fig.17.5(c) is an equilateral triangle. This fig. is reproduced in 17.6(a) below (right click and select 'open in new tab' for an enlarged view).
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| Fig.17.6 |
1. In the fig.17.6(a), two new
lines OA and PA are drawn. This gives us a new ΔOAP
2. Concentrate on this new
triangle.
3. OA = OP (∵ radii of the
same circle)
4. Now, 'A' lies on AB, which
is the perpendicular bisector of OP
5. Any point which lies on the
perpendicular bisector of a line will be equidistant from both the
ends of the line.
6. That is., any point on AB
will be equidistant from O and P. So A is equidistant from O and P
7. So PA = OA. But from (3) we
have OA = OP
8. Thus we get OA = OP = PA. Thus
ΔOAP is an equilateral triangle
9. we have some work to do
with this new equilateral ΔOAP. So it is shown separately as fig.17.5(b).
10. In the fig.b, the angle at
all the 3 vertices are marked as 60o. This is so because
it is an equilateral triangle
11. At the vertex A, a smaller
angle ∠OAM is marked as 30o. Why is this so?
12. We obtain ∠OAM as 30o,
by considering the inner ⊿OAM. In this right angled triangle, we have two
known angles:
13. ∠AMO = 90o
and ∠AOM = 60o. So ∠OAM = 180 – (90 + 60) = 180 –
150 = 30o.
14. Two results that we
obtained above are very important. They are: ∠OAM = 30o and
∠AOM = 60o. So let us mark them back on the fig.17.6(a). The
modified fig. is 17.6(c)
15. Now, in this fig.c, consider ∠AOM and
∠AOC. They are collinear. So ∠AOC = 180 – 60 = 120o.
16. Now consider ΔAOC. In
this triangle, Sides AO = CO (∵ radii of the same circle)
17. So ΔAOC is an isosceles
triangle. In any isosceles triangle, base angles are equal.
18. So we have: ∠CAO = ∠ACO.
As both are equal, let ∠CAO = ∠ACO = x
19. Then in ΔACO, x + x + 120
= 180o (∵ sum of interior angles in any Δ is180o)
20. From this we get x = 30o.
That is., ∠CAO = ∠ACO = x = 30o.
21. In the above, ∠AOC =
120o (15) and ∠CAO = 30o (20) are important
results. So let us mark it. The modified fig. is 17.6(d)
22. From fig.d, it is clear
that ∠CAB = 30 + 30 = 60o.
23. Also on fig.d, a new ΔOBP
is drawn. It resembles ΔOAP. We can do all the calculations that we
did on ΔOAP, on this new ΔOBP also.
24. OB = OP (∵ radii of the
same circle) [This step is similar to step(3)]
25. Now, 'B' lies on AB, which
is the perpendicular bisector of OP [This step is similar to step(4)]
26. Any point which lies on
the perpendicular bisector of a line will be equidistant from both
the ends of the line. [This step is similar to step(5)]
27. That is., any point on AB
will be equidistant from O and P. So B is equidistant from O and P.
[This step is similar to step(6)]
28. So PA = OB. But from (24)
we have OB = OP. [This step is similar to step(7)]
29. Thus we get OB = OP = PB.
Thus ΔOPB is an equilateral triangle [This step is similar to
step(8)]
30. As ΔPOB is equilateral,
all it’s angles are 60o. Thus we get ∠POB = 60o.
This is an important result. So it is marked in fig.d. From the fig.,
we get ∠AOB = 60 + 60 = 120o.
31. Consider the two
triangles: ΔOAC and ΔOAB.
32. OA is common
33. OB = OC (∵ radii of the
same circle)
34. Angle between these two
pairs of equal sides is also equal. It is 120o.
35. So it is a case of SAScongruence. (Side Angle Side). The two triangles are equal.
36. Thus we get AC = AB
37. The calculation steps are
complete. From the steps, we will pick two results:
♦ Result in (22): ∠CAB =
30 + 30 = 60o.
♦ Result in (36): AC = AB
• We have two equal sides AC and
AB, and their included angle = 60o. By theorem 17.2 that we saw above in this section, ΔABC
is an equilateral triangle.
• So we proved it. The chord AB,
which is the perpendicular bisector of a radius OP, is indeed one
side of the required equilateral triangle.
■ Thus we get an easy method to
draw an equilateral triangle in any given circle:
• First draw any convenient radius.
• Then draw it's perpendicular bisector.
• Extent this bisector both ways to make it a chord.
• This chord is one side of the required equilateral triangle.
• Extend the radius to meet the circle.
• This meeting point is the apex of the equilateral triangle.
• Join the apex to the base points. The equilateral triangle is complete
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