Chapter 32.9 - Excircle of a triangle

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.

Fig.32.58

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)

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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

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

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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

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

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We have seen that the three sides of a triangle touches the incircle. Now consider fig.32.61(a) below:

Fig.32.61

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 

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Let us take another set of two lines:
1. The same ΔABC is shown in the fig.32.62(a) below:

Fig.32.62

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. 

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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

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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

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:

Fig.32.64

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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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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.

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Chapter 32.8 - Incircle of a Triangle

In the previous section we saw the multiplication between a tangent and a chord. We also saw some solved examples. In this section we will see some details about incircles.

We have seen that, if we are given a circle, we can draw two tangents from any exterior point of the circle. See fig.32.38(b). It is shown again below:

Fig.32.38(b)

• We now ask a reverse question:
Given two lines which meet at a point. Can we draw a circle such that, both the lines are tangents to the circle?
Let us analyse this situation:
1. Consider fig.32.53(a) shown below. The two red lines AB and AC meet at A

Fig.32.53

• We want a circle in such a way that AB and AC are it's tangents.
• The yellow circle with center at O₁ is indeed such a circle
    ♦ This is because, the green radial lines from O₁ are perpendicular to both AB and AC at the respective points of contact. So AB and AC are tangents
• Is any other circle possible in this way?
2. Consider the cyan circle (with center at O₂) in fig.b. 
• AB and AC are tangents to this circle also
    ♦ This is because, the green radial lines from O₂ are perpendicular to both AB and AC at the respective points of contact. So AB and AC are tangents
• Is any other circle possible in this way?
3. Consider the magenta circle (with center at O₃) in fig.c. 
• AB and AC are tangents to this circle also
    ♦ This is because, the green radial lines from O₃ are perpendicular to both AB and AC at the respective points of contact. So AB and AC are tangents
4. It is clear that, a large number of such circles are possible. So what is the criteria?
• To find the answer, we want to know the property which is common to all such circles. 
• We can see that, the centres O₁, O₂, O₃ etc., of all such circles lie on the 'bisector of the angle between AB and AC'. 
• We can mark any point on the angle bisector. 
    ♦ Draw a perpendicular to AB through that point. 
    ♦ With the point as center, and the perpendicular distance as radius, draw a circle. 
    ♦ Then AB and AC will be tangents to the circle.
• We can mark infinite number of points on the angle bisector. Each of those point will give a circle.
• So it is clear that infinite number of such circles are possible

---

• Now we ask another question:
Given three lines which meet at three different points to give a triangle. Can we draw a circle such that, all the three lines are tangents to the circle?
Let us analyse this situation:
1. Consider fig.32.54(a) shown below. 

Fig.32.54

• The three red lines meet at A, B and C
• A blue circle is shown inside ΔABC. 
    ♦ Radial lines from the center O are perpendicular to the red lines at the respective points of contact. 
    ♦ So the blue circle is indeed our required circle. But how do we draw it? Let us try:
2. As before, we draw the 'bisector of the angle between AB and AC'. 
• We try different points O₁, O₂, O₃ etc., on that bisector. 
• We can draw circles using all those points as center. 
• All those circles will be tangential to AB and AC. 
• But there is no guarantee that, the circle draw in this way will be tangential to BC also
3. Next we try another angle bisector. This time the one between AB and BC. 
• The circles drawn are shown in fig.c. 
• All those circles will be tangential to AB and BC. 
• But there is no guarantee that, the circle draw in this way will be tangential to AC also
■ Thus we get an interesting result:
    ♦ The circles draw on the bisector between AB and AC are tangential to AB and AC
    ♦ The circles draw on the bisector between AB and BC are tangential to AB and BC
4. So consider the point of intersection of the two angle bisectors. It is marked as O in fig.32.55(a) below:

Fig.32.55

• From O, draw a perpendicular to AB. 
• With O as center and the perpendicular distance as radius, draw a circle. This is shown in fig.32.55(b). 
• All the lines AB, BC and AC will be tangential to the circle.
• This circle is called the incircle or inscribed circle of the triangle 
5. Interestingly, the 'bisector of the angle between AC and BC' will also pass through the same point O. We can write this:
■ The bisectors of all the three angles of a triangle meet at a point

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• Every triangle will have a circumcircle and incircle. See fig.32.56(a) below:

Fig.32.56

 • But for quadrilaterals, it depends on the type:
    ♦ Some quadrilaterals have both circumcircle and incircle. Example: Square. Fig.32.56(b) 
    ♦ Some quadrilaterals have only circumcircle but no incircle. Example: Rectangle. Fig.(c) 
    ♦ Some quadrilaterals have only incircle but no circumcircle. Example: Rhombus. Fig.(d) 
    ♦ Some quadrilaterals have neither circumcircle nor incircle. Example: Parallelogram. Fig.(e)

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• While making engineering drawings, 'tangents of circles' have much significance. 
• The fig.32.57 below shows a part of a screen shot of a cad program while drawing circles. 

Fig.32.57

■ The red arrow shows the option 'Ttr'. That is., 'Tangent, tangent, radius'. 
• If we give two tangents only, infinite number of circles are possible. 
• This is clear from fig.32.53(c) that we saw above. 
• So we must specify a radius also.
■ The green arrow shows the option 'TTT'. That is., 'Tangent, Tangent, Tangent'. 
• If we give three tangents, then the unique circle can be clearly specified. 
• This is clear from fig.32.55(b) above. 
• Additional information about radius is not required.

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In the next section, we will see the relation between the incircle and the sides of a triangle.

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Chapter 32.7 - Multiplication of Tangent and Chord

In the previous section we saw the two tangents from an exterior point. We also saw some solved examples. In this section we will see some more advanced details related to tangents.

We have seen the case when two chords of a circle intersect outside the circle. See fig.27.68 in chapter 27. That fig. is shown again below:

Fig.27.68

Based on the fig., we derived the following result:
PA × PB = PC × PD
■ But what if one of them is a tangent?
Such a situation is shown in fig.32.46(a) below:

Fig.32.46

Let us do an analysis. We will write the steps:
1. One line from P cuts the circle at A and B
• The other line from P is a tangent, which touches the circle at C
2. We want the relation between these two line. 
• For that, draw the chords AC and BC. This is shown in fig.b
• Now, ∠PCA is the angle between the chord AC and the tangent on the 'tangent side'
• ∠ABC is the angle made by the chord AC (at a point on the circle) on the non-tangent side
• Those two angles will be the same. (Theorem 32.7)
3. Separate the two triangles ΔPAC and ΔPBC as shown in fig.c
• Angle at P is the same in both triangles
• Angle at C in ΔPAC = Angle at B in ΔPBC
■ Two angles are same. So the third angle will also be the same. That is., 
• Angle at A in ΔPAC = Angle at C in ΔPBC
4. So all the three angles are the same. 
The two triangles ΔPAC and ΔPBC are similar. 
5. Now we apply a special property that is applicable to any two similar triangles (Details here):

^{side opposite smallest angle in ΔPAC}⁄_{side opposite smallest angle in ΔPBC} 

= ^{side opposite medium angle in ΔPAC}⁄_{side opposite medium angle in ΔPBC}

= ^{side opposite largest angle in ΔPAC}⁄_{side opposite largest angle in ΔPBC} 

5. But angles in the two triangles are the same. That is.,

• Smallest angle in ΔPAC = Smallest angle in ΔPBC

• Medium angle in ΔPAC = Medium angle in ΔPBC

• Largest angle in ΔPAC = Largest angle in ΔPBC

6. So we can write this:

• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,

^{side opposite P in ΔPAC}⁄_{side opposite P in ΔPBC}
= ^{side opposite A in ΔPAC}⁄_{side opposite C in ΔPBC}
= ^{side opposite C in ΔPAC}⁄_{side opposite B in ΔPBC}
7. So we get: ^AC⁄_BC = ^PC⁄_PB = ^PA⁄_PC 

8. Take the second and the third ratios. We get:
^• PC⁄_PB = ^PA⁄_PC . From this we get:
PA × PB = PC².
This is a very useful result. We can write it as a theorem.
Theorem 32.9:
1. Two lines are drawn from an exterior point P towards a circle
• One of them cuts the circle at two points A and B
• The other is a tangent which touches the circle at C
2. Consider the line which cuts the circle
(i) It's whole length is PB
(ii) It's length outside the circle is PA
3. Take the product. That is:
(Whole length) × (Length outside)
■ This product will be equal to the square of the length of the tangent

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Let us see some solved examples:
Solved example 32.21
In the fig.32.47(a) below, AB is a diameter and P is a point on AB extended. 

Fig.32.47

A tangent from P touches the circle at Q. What is the radius of the circle?
Solution:
1. We can apply the above theorem 32.9. In this case we can write:
• Two lines are drawn from an exterior point P towards a circle
• One of them cuts the circle at two points A and B
• The other is a tangent which touches the circle at Q
2. So we can write: PA × PB = PQ²
⟹ 8 × PB = 4² ⟹ PB = ¹⁶⁄₈ = 2 cm
3. Thus we get: AB = AP-PB = 8-2 = 6 cm
• So radius = ⁶⁄₂ = 3cm 

Solved example 32.22
In the fig.32.47(b) above, the line joining two points A and B on a circle is extended by 4 cm to reach P and a tangent is drawn from P. Length of that tangent PQ is 6 cm.
In the fig.32.47(c), P is further extended along the same line AB by 1 cm. Then a tangent PR is drawn. What is the length of the tangent PR?
Solution:
1. Applying theorem 32.9 to fig.b, we get:
PA × PB = PQ²  ⟹ PA × 4 = 6²  ⟹ PA = ³⁶⁄₄ = 9 cm
• So we get PA in fig.b. We can use it in fig.c. 
But since there is a further extension of 1 cm, the new PA in fig.c is (9+1) = 10 cm
2. Applying theorem 32.9 to fig.c, we get:
PA × PB = PR² ⟹ 10 × 5 = PR² ⟹ PR = √50 = √[2×25] = √2 × √25 = 5√2 cm

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Let us see a practical application of theorem 32.9:
• We derived theorem 32.9 based on fig.32.46(a) above. The same fig. is shown again as fig.32.48(a) below:

`Fig.32.48

1. Consider the segment PC. Let it be the side of a square. 
• If we are given any one side of a square, we can easily construct the whole square. The completed square is named as PQRC in fig.b
• Area of PQRC is PC².
2. Draw a line perpendicular to PB
• With P as center and PB as radius, draw an arc cutting the perpendicular line at S. Then PS = PB
3. Consider PA and PS as the two adjacent sides of a rectangle. 
• Knowing the two adjacent sides, we can easily complete the whole rectangle.
• It is named as PSTA in fig.c
• Area of rectangle PSTA = PA × PS = PA × PB (∵ PS = PB)
4. But from theorem 32.9, we have: PA × PB = PC².
So we can write:
■ Area of rectangle PSTA = Area of square PQRC
• This gives us an effective method to draw a square which has the same area as a given rectangle and vice versa

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Let us see some solved example:
Solved example 32.23
Draw a rectangle of one side 6 cm and area equal to that of a square of side 5 cm
Solution:
Step 1 (fig.32.49.a):

Fig.32.49

1. Draw a horizontal line AB, 6 cm in length.
2. Draw a perpendicular bisector of AB and thus mark the center O₁ of AB
3. With O₁ as center, draw a semicircle with AB as diameter
4. With A as center, draw an arc of radius 5 cm, cutting the semicircle at C. 
• Then ∠ACB will be 90^o. (Details here)
Step 2 (fig.32.49.b):
1. Draw a perpendicular bisector of BC and thus mark the center O₂ of BC
2. With O₂ as center, draw a circle with BC as diameter
3. Mark the point of intersection of this circle with AB as D
Step 3 (fig.32.49.c):
1. Draw a perpendicular to AB
2. With A as center and AB as radius, draw an arc intersecting the perpendicular at E
3. Complete the rectangle AEFD by considering AD and AE is two adjacent sides
Explanation:
1. From theorem 32.9, we get: 
AD × AB = AC² ⟹ AD × AE = AC² ⟹ AD × AE = 5² 
2. So Area of rectangle AEFD is same as that of a square of side 5 cm. 
3. Thus AEFD is the required rectangle.

Solved example 32.24
Draw a square of area 25 sq.cm. Draw a rectangle of one side 8 cm and area equal to that of the square.
Solution:
Step 1 (fig.32.50.a):

Fig.32.50

1. Draw a horizontal line AB, 8 cm in length.
2. Draw a perpendicular bisector of AB and thus mark the center O₁ of AB
3. With O₁ as center, draw a semicircle with AB as diameter
4. With A as center, draw an arc of radius 5 cm, cutting the semicircle at C. 
• Then ∠ACB will be 90^o. (Details here)
Step 2 (fig.32.50.b):
1. Draw a perpendicular bisector of BC and thus mark the center O₂ of BC
2. With O₂ as center, draw a circle with BC as diameter
3. Mark the point of intersection of this circle with AB as D
Step 3 (fig.32.50.c):
1. Draw a perpendicular to AB
2. With A as center and AB as radius, draw an arc intersecting the perpendicular at E
3. Complete the rectangle AEFD by considering AD and AE is two adjacent sides
Explanation:
1. From theorem 32.9, we get: 
AD × AB = AC² ⟹ AD × AE = AC² ⟹ AD × AE = 5² 
2. So Area of rectangle AEFD is same as that of a square of side 5 cm. 
3. Thus AEFD is the required rectangle

Solved example 32.25
Draw a rectangle of one side 6 cm and area equal to that of a square of side 4 cm
Solution:
Step 1 (fig.32.51.a):

Fig.32.51

1. Draw a horizontal line AB, 6 cm in length.
2. Draw a perpendicular bisector of AB and thus mark the center O₁ of AB
3. With O₁ as center, draw a semicircle with AB as diameter
4. With A as center, draw an arc of radius 4 cm, cutting the semicircle at C. 
• Then ∠ACB will be 90^o. (Details here)
Step 2 (fig.32.51.b):
1. Draw a perpendicular bisector of BC and thus mark the center O₂ of BC
2. With O₂ as center, draw a circle with BC as diameter
3. Mark the point of intersection of this circle with AB as D
Step 3 (fig.32.51.c):
1. Draw a perpendicular to AB
2. With A as center and AB as radius, draw an arc intersecting the perpendicular at E
3. Complete the rectangle AEFD by considering AD and AE is two adjacent sides
Explanation:
1. From theorem 32.9, we get: 
AD × AB = AC² ⟹ AD × AE = AC² ⟹ AD × AE = 4² 
2. So Area of rectangle AEFD is same as that of a square of side 4 cm. 
3. Thus AEFD is the required rectangle.

Solved example 32.26
Draw a rectangle of one side 7 cm and area 36 sq.cm
Solution:
Step 1 (fig.32.52.a):

Fig.32.52

1. Draw a horizontal line AB, 7 cm in length.
2. Draw a perpendicular bisector of AB and thus mark the center O₁ of AB
3. With O₁ as center, draw a semicircle with AB as diameter
4. With A as center, draw an arc of radius 6 cm, cutting the semicircle at C. 
• Then ∠ACB will be 90^o. (Details here)
Step 2 (fig.32.52.b):
1. Draw a perpendicular bisector of BC and thus mark the center O₂ of BC
2. With O₂ as center, draw a circle with BC as diameter
3. Mark the point of intersection of this circle with AB as D
Step 3 (fig.32.52.c):
1. Draw a perpendicular to AB
2. With A as center and AB as radius, draw an arc intersecting the perpendicular at E
3. Complete the rectangle AEFD by considering AD and AE is two adjacent sides
Explanation:
1. From theorem 32.9, we get: 
AD × AB = AC² ⟹ AD × AE = AC² ⟹ AD × AE = 6² 
2. So Area of rectangle AEFD is same as that of a square of side 6 cm. 
3. Thus AEFD is the required rectangle.

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In the next section, we will see more details about Tangents.

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Chapter 32.6 - Two tangents - Solved examples

In the previous section we saw the two tangents from an exterior point. We also saw a solved example. In this section we will see a few more solved examples.

Solved example 32.17
In fig.32.43(a) below, ABC is an isosceles triangle. AC and BC are the equal sides and AB is the base. 

Fig.32.43

All the three sides are tangents to the circle and the tangent points are P, Q and R. Prove that the base AB is bisected at P
Solution:
1. Tangents through P and Q meet at B. So BP = BQ
2. Tangents through Q and R meet at C. So CR = CQ
3. Tangents through P and R meet at A. So AP = AR
4. Given that AC = BC. So we can write:
AR + CR = BQ + CQ
5. But from (2), we have CR = CQ
So (4) becomes: AR = BQ
6. But from (3), we have: AR = AP
Also from (1), we have: BQ = BP
■ So (5) becomes: AP = BP. That means, AB is bisected at P

Solved example 32.18
In fig.32.43(b) above, a circle is touching the side AB of a ΔABC at Q. The circle is also touching the side CA produced at P and CB produced at R. Prove that: CP = Half of the perimeter of ΔABC
Solution:
1. (i) Tangents through P and Q meet at A. So AP = AQ
(ii) Tangents through Q and R meet at B. So BQ = BR
(iii) Tangents through P and R meet at C. So CP = CR
2. The perimeter of ΔABC = AB + BC + CA
• But AB = AQ + BQ
• So perimeter = AQ + BQ + BC + CA
3. From 1(i), we have: AQ = AP
• Also from 1(ii), we have: BQ = BR
• So (2) becomes:
Perimeter = AP + BR + BC + CA
4. Consider the first and last terms in the above equation:
• They are AP and CA
• From the fig., we get: AP + CA = CP
5. So (3) becomes:
• Perimeter =  CP + BR + BC
6. Also from the fig., BR + BC = CR
• So (5) becomes: 
Perimeter = CP + CR
7. But from 1(iii), we have: CP = CR
• So (6) becomes:
Perimeter = CP + CP = 2 CP
■ Thus we get: CP = Half of perimeter

Solved example 32.19
In fig.32.44(a) below, 3 tangents are drawn at P, Q and R on the circle. The 3 tangents intersect at 3 points to form ΔABC. The tangents at P and R are perpendicular to each other. 

Fig.32.44

Prove that the perimeter of ΔABC is equal to the diameter of the circle.
Solution:
1. From the solved example 32.18 above, we know that AP in fig.32.44(a) = Half the perimeter of ΔABC
2. Now draw the radial lines from R and P. This is shown in fig.b
• Since AR is a tangent, ∠ARO = 90^o 
• Since AP is a tangent, ∠APO = 90^o
3. So in the quadrilateral APOR, 3 angles are 90^o. They are:
∠ARO, ∠APO and ∠PAR
• So the fourth angle ∠POR will also be 90^o
4. In the quadrilateral APOR, all four angles are 90^o
• And adjacent sides AP = AR (∵ they are tangents from an exterior point A)
• So APOR is a square
5. In the square APOR, AP will be equal to OR. So we can write:
• AP = OR
⟹ Half the perimeter of ΔABC = OR
⟹ Full perimeter of ΔABC = 2 × Half perimeter of ΔABC = 2 × OR
But 2 × OR = 2 × radius = diameter of the circle. 
■ So we get:
Full perimeter of ΔABC = diameter of the circle

Solved example 32.20
In the fig.32.45(a) below, two circles touch at a point P and a common tangent AB is drawn there

Fig.32.45

(i) Prove that this common tangent AB bisects the segment RQ on another common tangent CD of the circles (fig.b). Where R and Q are the points of contact.
(ii) Prove that the points of contact P, Q and R form the vertices of a right triangle (fig.c)
Solution:
Part (i):
Let the two tangents AB and CD intersect at S
1. Tangents through P and Q meet at S. So SP = SQ

2. Tangents through P and R meet at S. So SP = SR
3. From (1) and (2) we get: SR = SQ
• So the point S bisects the segment RQ. 
■ In other words, the tangent AB bisects the segment RQ
Part (ii):
Consider fig.32.45(d) below. 

Fig.32.45(d)

1. If we draw an arc with S as center and SR as radius, the points P and Q will lie on that arc.
This is because SR = SP = SQ
2. Since R and S lie on the same line, the arc RPQ is a semicircle and RQ is the diameter of that semicircle
3. So for the ∠RPQ, R and Q are the ends of the diameter and P lies on the semicircle. Thus ∠RPQ is a right angle

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In the next section, we will see more details about Tangents.

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