In the previous section we saw the details about incircles. In this section we will see the relation between incircle and the sides of a triangle.
1. Consider the ΔABC and it's incircle in fig.32.58(a) below. P, Q and R are the tangent points.
We can write the following 3 points:
(i) From the point A, two tangents AP and AR are drawn
So AP and AR are equal in length [Theorem 32.2]
Let AP = AR = x
(ii) From the point B, two tangents BP and BQ are drawn
So BP and BQ are equal in length
Let BP = BQ = y
(iii) From the point C, two tangents CQ and CR are drawn
So CQ and CR are equal in length
Let CQ = CR = z
2. Perimeter of ΔABC = p = AB + BC + AC = (x+y) + (y+z) + (x+z) = 2x + 2y + 2z
⟹ p = 2(x+y+z)
• So if s = half of the perimeter, we will get:
s = p⁄2 = (x+y+z)
3. Now consider fig.32.58(b). It shows the same ΔABC in fig.a
• But the lengths of sides are shown in a different manner:
♦ Length of the side opposite vertex A is a
♦ Length of the side opposite vertex B is b
♦ Length of the side opposite vertex C is c
• So p = (a+b+c)
4. Combining figs. (a) and (b), we get:
• x+y = c
• y+z = a
• x+z = b
5. Then we will get:
• x = (x+y+z) - (y+z) = (s-a)
• y = (x+y+z) - (x+z) = (s-b)
• z = (x+y+z) - (x+y) = (s-c)
• These 3 distances are shown in fig.c.
• From the fig., we can see a pattern:
♦ From vertex A, the two distances are: (s-a) and (s-a)
♦ From vertex B, the two distances are: (s-b) and (s-b)
♦ From vertex C, the two distances are: (s-c) and (s-c)
Let us see an example:
• In the fig.32.59(a) below, ABC has sides 5 cm, 6 cm and 7 cm
1. Perimeter p = (5+6+7) = 18 cm
So semi perimeter s = 18⁄2 = 9 cm
2. As mentioned in the above discussion,
• Length of the side opposite vertex A should be written as a
• Length of the side opposite vertex B should be written as b
• Length of the side opposite vertex C should be written as c
This is shown in fig.b
• So we get:
♦ The two distances from vertex A are: (s-a) = (9-7) = 2 cm
♦ The two distances from vertex B are: (s-b) = (9-6) = 3 cm
♦ The two distances from vertex C are: (s-c) = (9-5) = 4 cm
This is shown in fig.c
■ So, even with out drawing the incircle, or knowing it's radius, we are able to mark the points where the incircle split the sides
Let us now see the area of the triangle:
1. Consider the ΔABC and it's incircle in fig.32.60(a) below. It is the same one that we saw in fig.32.58(a) above.
2. It is shown again in fig32.60(b). But this time three new magenta lines are drawn. Each one from a vertex to the centre O of the incircle.
3. So we get three triangles: ΔOAB, ΔOBC and ΔOCA
• We want the area of each of these three triangles. Let us try to find them:
4. Consider ΔOAB. We know that the area of a triangle is: 1⁄2 × base × altitude
• For OAB, the base can be taken as AB.
• Then the altitude is already present. It is the radial line OP (∵ OP is perpendicular to AB)
• So Area of OAB = 1⁄2 × AB × OP = 1⁄2 × c × r = cr⁄2 (where PO = r, the radius of the incircle)
• Thus Area of ΔOAB = cr⁄2
5. In the same way, we will get:
• Area of OBC = ar⁄2
• Area of OCA = br⁄2
6. So sum of the three areas = (ar⁄2 + br⁄2 + cr⁄2) = (r(a+b+c)⁄2) = rs [∵ (a+b+c)⁄2 = s, the semi perimeter]
7. But the sum of the three areas is the total area of the ΔABC. If we put the total area of ABC as 'A', we can write:
• A = rs ⟹ r = A⁄s
We can write the above result as a theorem:
Theorem 32.10:
If we have the area of a triangle and the length of all it's three sides, we can find the radius of it's incircle using the equation: r = A⁄s.
• Where s is the semi perimeter
We have seen that the three sides of a triangle touches the incircle. Now consider fig.32.61(a) below:
1. The original triangle is ΔABC. It's side AB is extended towards the left up to A'.
• Then ∠A'AC is an external angle of ΔABC
• The bisector of ∠A'AC is drawn in dashed red line
2. Mark any point O1 on this angle bisector.
3. Draw a perpendicular (shown in green lines in fig.b) from O1 onto either AC or AA'
4. Draw a circle with that perpendicular length as radius.
• We will get a circle which touches AA' and AC.
• Note that, the circle does not touch the side BC
5. We can mark any number of points like O1 on the angle bisector and draw circles in this manner. All those circles will touch AA' and AC. But BC will not be touched
■ AA' is actually, an extension of the side AB. So we can write:
The circles touches two sides AB and AC. The side BC is left out
Let us take another set of two lines:
1. The same ΔABC is shown in the fig.32.62(a) below:
It's side BC is extended towards the top up to C'.
• Then ∠C'CA is an external angle of ΔABC
• The bisector of ∠C'CA is drawn in dashed red line
2. Mark any point O2 on this angle bisector.
3. Draw a perpendicular (shown in green lines in fig.b) from O2 onto either AC or CC'
4. Draw a circle with that perpendicular length as radius.
• We will get a circle which touches CC' and AC.
• Note that, the circle does not touch the side AB
5. We can mark any number of points like O2 on the angle bisector and draw circles in this manner. All those circles will touch CC' and AC. But AB will not be touched.
■ CC' is actually, an extension of the side BC. So we can write:
The circles touches two sides BC and AC. The side AB is left out.
Let us write a summary. We will write it in steps:
1. Consider any one of the three vertices of a triangle
2. Draw the exterior angle at that corner
3. Draw the angle bisector of that exterior angle
4. Mark any point on that angle bisector
5. Draw the circle which touches the sides of the triangle
6. That circle will touch only those two sides which meet at the vertex taken in (i) above.
7. The third side will be left out
So what do we do to make the circle touch the third side also?
Ans: The same as we did for incircle:
■ Draw the circle using the 'point of intersection of any two angle bisectors' as the center.
Let us see how it is done:
1. In fig.32.63(a) below, two angle bisectors are drawn:
• Angle bisector of ∠A'AC
• Angle bisector of ∠C'CA
• The two angle bisectors intersect at O'
3. Draw a perpendicular from O' up to any one of the three lines: AA', AC or CC'
• This is shown in green colour in fig.b
4. With O' as center and the perpendicular distance as radius, draw a circle. That circle will touch all the three sides of the triangle
■ Such a circle is called the excircle or escribed circle of the triangle
Three excircles are possible for any triangle. For our ΔABC, the other two are shown in fig.32.64 below:
Note the following points:
• The angle bisectors of ∠ACC' and ∠BCC'' are the same
• The angle bisectors of ∠CBB' and ∠ABB'' are the same
• The angle bisectors of ∠CAA' and ∠BAA'' are the same
• The angle bisectors intersect at three points to form a triangle
These points, though interesting, are not relevant for our discussion in this chapter.
To continue our present discussion, we need the incircle and any one of the three excircles in fig.32.64 above. We will choose the cyan circle. We will see them in the next section.
1. Consider the ΔABC and it's incircle in fig.32.58(a) below. P, Q and R are the tangent points.
 |
| Fig.32.58 |
(i) From the point A, two tangents AP and AR are drawn
So AP and AR are equal in length [Theorem 32.2]
Let AP = AR = x
(ii) From the point B, two tangents BP and BQ are drawn
So BP and BQ are equal in length
Let BP = BQ = y
(iii) From the point C, two tangents CQ and CR are drawn
So CQ and CR are equal in length
Let CQ = CR = z
2. Perimeter of ΔABC = p = AB + BC + AC = (x+y) + (y+z) + (x+z) = 2x + 2y + 2z
⟹ p = 2(x+y+z)
• So if s = half of the perimeter, we will get:
s = p⁄2 = (x+y+z)
3. Now consider fig.32.58(b). It shows the same ΔABC in fig.a
• But the lengths of sides are shown in a different manner:
♦ Length of the side opposite vertex A is a
♦ Length of the side opposite vertex B is b
♦ Length of the side opposite vertex C is c
• So p = (a+b+c)
4. Combining figs. (a) and (b), we get:
• x+y = c
• y+z = a
• x+z = b
5. Then we will get:
• x = (x+y+z) - (y+z) = (s-a)
• y = (x+y+z) - (x+z) = (s-b)
• z = (x+y+z) - (x+y) = (s-c)
• These 3 distances are shown in fig.c.
• From the fig., we can see a pattern:
♦ From vertex A, the two distances are: (s-a) and (s-a)
♦ From vertex B, the two distances are: (s-b) and (s-b)
♦ From vertex C, the two distances are: (s-c) and (s-c)
Let us see an example:
• In the fig.32.59(a) below, ABC has sides 5 cm, 6 cm and 7 cm
 |
| Fig.32.59 |
So semi perimeter s = 18⁄2 = 9 cm
2. As mentioned in the above discussion,
• Length of the side opposite vertex A should be written as a
• Length of the side opposite vertex B should be written as b
• Length of the side opposite vertex C should be written as c
This is shown in fig.b
• So we get:
♦ The two distances from vertex A are: (s-a) = (9-7) = 2 cm
♦ The two distances from vertex B are: (s-b) = (9-6) = 3 cm
♦ The two distances from vertex C are: (s-c) = (9-5) = 4 cm
This is shown in fig.c
■ So, even with out drawing the incircle, or knowing it's radius, we are able to mark the points where the incircle split the sides
Let us now see the area of the triangle:
1. Consider the ΔABC and it's incircle in fig.32.60(a) below. It is the same one that we saw in fig.32.58(a) above.
 |
| Fig.32.60 |
3. So we get three triangles: ΔOAB, ΔOBC and ΔOCA
• We want the area of each of these three triangles. Let us try to find them:
4. Consider ΔOAB. We know that the area of a triangle is: 1⁄2 × base × altitude
• For OAB, the base can be taken as AB.
• Then the altitude is already present. It is the radial line OP (∵ OP is perpendicular to AB)
• So Area of OAB = 1⁄2 × AB × OP = 1⁄2 × c × r = cr⁄2 (where PO = r, the radius of the incircle)
• Thus Area of ΔOAB = cr⁄2
5. In the same way, we will get:
• Area of OBC = ar⁄2
• Area of OCA = br⁄2
6. So sum of the three areas = (ar⁄2 + br⁄2 + cr⁄2) = (r(a+b+c)⁄2) = rs [∵ (a+b+c)⁄2 = s, the semi perimeter]
7. But the sum of the three areas is the total area of the ΔABC. If we put the total area of ABC as 'A', we can write:
• A = rs ⟹ r = A⁄s
We can write the above result as a theorem:
Theorem 32.10:
If we have the area of a triangle and the length of all it's three sides, we can find the radius of it's incircle using the equation: r = A⁄s.
• Where s is the semi perimeter
We have seen that the three sides of a triangle touches the incircle. Now consider fig.32.61(a) below:
 |
| Fig.32.61 |
• Then ∠A'AC is an external angle of ΔABC
• The bisector of ∠A'AC is drawn in dashed red line
2. Mark any point O1 on this angle bisector.
3. Draw a perpendicular (shown in green lines in fig.b) from O1 onto either AC or AA'
4. Draw a circle with that perpendicular length as radius.
• We will get a circle which touches AA' and AC.
• Note that, the circle does not touch the side BC
5. We can mark any number of points like O1 on the angle bisector and draw circles in this manner. All those circles will touch AA' and AC. But BC will not be touched
■ AA' is actually, an extension of the side AB. So we can write:
The circles touches two sides AB and AC. The side BC is left out
Let us take another set of two lines:
1. The same ΔABC is shown in the fig.32.62(a) below:
 |
| Fig.32.62 |
• Then ∠C'CA is an external angle of ΔABC
• The bisector of ∠C'CA is drawn in dashed red line
2. Mark any point O2 on this angle bisector.
3. Draw a perpendicular (shown in green lines in fig.b) from O2 onto either AC or CC'
4. Draw a circle with that perpendicular length as radius.
• We will get a circle which touches CC' and AC.
• Note that, the circle does not touch the side AB
5. We can mark any number of points like O2 on the angle bisector and draw circles in this manner. All those circles will touch CC' and AC. But AB will not be touched.
■ CC' is actually, an extension of the side BC. So we can write:
The circles touches two sides BC and AC. The side AB is left out.
Let us write a summary. We will write it in steps:
1. Consider any one of the three vertices of a triangle
2. Draw the exterior angle at that corner
3. Draw the angle bisector of that exterior angle
4. Mark any point on that angle bisector
5. Draw the circle which touches the sides of the triangle
6. That circle will touch only those two sides which meet at the vertex taken in (i) above.
7. The third side will be left out
So what do we do to make the circle touch the third side also?
Ans: The same as we did for incircle:
■ Draw the circle using the 'point of intersection of any two angle bisectors' as the center.
Let us see how it is done:
 |
| Fig.32.63 |
• Angle bisector of ∠A'AC
• Angle bisector of ∠C'CA
• The two angle bisectors intersect at O'
3. Draw a perpendicular from O' up to any one of the three lines: AA', AC or CC'
• This is shown in green colour in fig.b
4. With O' as center and the perpendicular distance as radius, draw a circle. That circle will touch all the three sides of the triangle
■ Such a circle is called the excircle or escribed circle of the triangle
Three excircles are possible for any triangle. For our ΔABC, the other two are shown in fig.32.64 below:
 |
| Fig.32.64 |
Note the following points:
• The angle bisectors of ∠ACC' and ∠BCC'' are the same
• The angle bisectors of ∠CBB' and ∠ABB'' are the same
• The angle bisectors of ∠CAA' and ∠BAA'' are the same
• The angle bisectors intersect at three points to form a triangle
These points, though interesting, are not relevant for our discussion in this chapter.
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