In the previous section we saw some tangents from an exterior point. In this section we will learn more details.
1. Consider the two tangents AP and BP in fig.32.15 below.
• OA and OB are radii.
♦ The tangents are drawn at the points of intersection of the radii with the circle
2. We know that ∠OAP = ∠OBP = 90o
3. Now consider the quadrilateral OAPB. It is formed by the radii and the tangents
4. A and B are two opposite corners. The sum of the angles at those opposite corners = 90+90 = 180o
■ If the sum of opposite angles in any quadrilateral is 180o, that quadrilateral will be a cyclic quadrilateral,
• That is., all the four vertices of that quadrilateral will lie on a circle (We saw details here)
5. So we can draw a circle through all the four vertices of the quadrilateral OAPB. This circle is shown in green colour in fig.32.15(b) above
We can write the above result in the form of a theorem:
Theorem 32.4
• If a quadrilateral has the following four vertices:
♦ First vertex at the centre of a circle
♦ Second vertex at the exterior point from which two tangents are drawn to the circle
♦ Third vertex at one tangent point
♦ Fourth vertex at the other tangent point
■ Then that quadrilateral is cyclic
From the above, we get another useful result:
• In the fig.32.15(b) above, we considered the opposite vertices A and B. What about the other two opposite vertices O and P?
• Obviously, the sum of those two angles must also be 180o (∵ the sum of interior angles of any quadrilaterals is 360o)
This result is also useful for solving problems. We will write it as a theorem:
Theorem 32.5
1. In a circle, two radii are drawn
2. Tangents are drawn at the end point of each of those radii
• These tangents meet at P
3. Then sum of the following two angles is 180o
• Angle at O between the two radii
• Angle at P between the two tangents
Now we will see an application of this theorem in the form of a solved example
Solved example 32.6
Draw an equilateral triangle in such a way that it's all three sides are tangential to a circle of radius 3 cm
Solution:
A rough sketch is shown in fig.32.16(a) below:
• The three sides of the triangle are: AB, BC and AC. They are all tangents to the circle of radius 3 cm
• The triangle is to be equilateral. So all it's angles are 60
• Based on this rough fig., we must obtain details to make the actual construction
We will write the steps:
1. From the center O, draw radial lines OP, OQ and OR. This is shown in fig.b
2. Based on theorem 32.5, we get:
• ∠A + ∠ROP = 180 ⟹ 60 + ∠ROP = 180 ⟹ ∠ROP = 120
• ∠B + ∠POQ = 180 ⟹ 60 + ∠POQ = 180 ⟹ ∠POQ = 120
• ∠C + ∠ROQ = 180 ⟹ 60 + ∠ROQ = 180 ⟹ ∠ROQ = 120
• So the three central angles are all 120o. This is marked in fig.c
• We can now begin the actual construction
3. Step 1 is to draw a circle of radius 3 cm. Mark the center as 'O'. Then draw a radial line OP in any convenient direction. This is shown in fig.32.17(a) below:
4. Step 2 is to draw the other two radial lines:
• Draw OQ in such a way that it makes an angle of 120o with OP
• Draw OR in such a way that it makes an angle of 120o with OQ
• Then the third angle ∠POR will be naturally 120
This is shown in fig.32.17(b)
5. Step 3 is the final step. In this step we draw the tangents:
• Draw a line perpendicular to OP at P. This is the tangent at P
• Draw a line perpendicular to OQ at Q. This is the tangent at Q
• Draw a line perpendicular to OR at R. This is the tangent at R
■ The three tangents will meet at A, B and C, giving the required equilateral triangle. This is shown in fig.c
Solved example 32.7
Draw a triangle of angles 40o, 60o and 80o in such a way that it's all three sides are tangential to a circle of radius 2.5 cm
Solution:
A rough sketch is shown in fig.32.18(a) below:
• The three sides of the triangle are: AB, BC and AC. They are all tangents to the circle of radius 2.5 cm
• The required angles are also marked in the fig.a
• Based on this rough fig., we must obtain details to make the actual construction
We will write the steps:
1. From the center O, draw radial lines OP, OQ and OR. This is shown in fig.b
2. Based on theorem 32.5, we get:
• ∠A + ∠ROP = 180 ⟹ 40 + ∠ROP = 180 ⟹ ∠ROP = 140
• ∠B + ∠POQ = 180 ⟹ 60 + ∠POQ = 180 ⟹ ∠POQ = 120
• ∠C + ∠ROQ = 180 ⟹ 80 + ∠ROQ = 180 ⟹ ∠ROQ = 100
• So the three central angles are 140o, 120o and 100o. This is marked in fig.c
• We can now begin the actual construction
3. Step 1 is to draw a circle of radius 2.5 cm. Mark the center as 'O'. Then draw a radial line OP in any convenient direction. This is shown in fig.32.19(a) below:
4. Step 2 is to draw the other two radial lines:
• Draw OQ in such a way that it makes an angle of 120o with OP
• Draw OR in such a way that it makes an angle of 100o with OQ
• Then the third angle ∠POR will be naturally 140o
This is shown in fig.32.19(b)
5. Step 3 is the final step. In this step we draw the tangents:
• Draw a line perpendicular to OP at P. This is the tangent at P
• Draw a line perpendicular to OQ at Q. This is the tangent at Q
• Draw a line perpendicular to OR at R. This is the tangent at R
■ The three tangents will meet at A, B and C, giving the required triangle. This is shown in fig.c
Solved example 32.8
In the fig.32.20(a), the blue triangle is equilateral. It's circumcircle is shown in green colour. All the sides of the red triangle are tangents to the green circle. The tangent points are the vertices of the blue triangle.
(i) Prove that the red triangle is also equilateral and it's sides are double that of the blue triangle
(ii) Draw the fig. with sides of the small triangle 3 cm
Solution:
Part 1:
Let ABC be the red triangle and PQR be the blue triangle. This is marked in fig.b
1. Mark the center O of the green circle. Draw radial lines OP, OQ and OR. They are shown in yellow colour
2. Given that the red lines are tangents at the vertices. So the yellow lines are perpendicular to the red lines. we can write:
AB is perpendicular to OP
BC is perpendicular to OQ
AC is perpendicular to OR
3. Each of the three blue lines PQ, QR and RP subtend an angle at the center O:
♦ PQ subtend ∠POQ
♦ QR subtend ∠QOR
♦ RP subtend ∠ROP
• But PQ = QR = RP (∵ they are sides of an equilateral triangle)
• So the angle subtended by them must be equal
♦ Thus we get: ∠POQ = ∠QOR = ∠ROP
• But the total angle at any point is 360o
♦ We can write: ∠POQ + ∠QOR + ∠ROP = 360o
• So, since the three angles are equal, we get: ∠POQ = ∠QOR = ∠ROP = 360⁄3 = 120o
4. Based on theorem 32.5, we can write:
• ∠A + ∠ROP = 180 ⟹ ∠A+ 120 = 180 ⟹ ∠A = 60
• ∠B + ∠POQ = 180 ⟹ ∠B+ 120 = 180 ⟹ ∠B = 60
• ∠C + ∠ROQ = 180 ⟹ ∠C+ 120 = 180 ⟹ ∠C = 60
• So the three angles of the red triangle are all 60o. This is marked in fig.b. Thus the red triangle ABC is an equilateral triangle.
5. Consider fig.c: AB and AC are two tangents from an exterior point A
(i) P and R are the tangent points. Then by theorem 32.2 that we saw in the previous section, AP = AR
(ii) In the same way, we have: BP = BQ and CQ = CR
(iii) Let: AP = AR = a
BP = BQ = b
CQ = CR = c
(iv) Since the triangle is equilateral, we have: AB = BC = CA ⟹ (a+b) = (b+c) = (c+a)
From (a+b) = (b+c), we get: a = c
From (b+c) = (c+a), we get: a = b
• So we have: a = b = c
6. P is the midpoint of AB (∵ a = b)
R is the midpoint of AC (∵ a = c)
• So the line PR cuts the sides AB and AC at their midpoints
• Then by theorem 18.5, (Division of triangles by parallel lines) we get:
♦ BC is parallel to PR
♦ BC is double the length of PR
In the same way:
♦ AC is parallel to PQ
♦ AC is double the length of PQ
AND
♦ AB is parallel to RQ
♦ AB is double the length of RQ
■ (4) and (6) together gives the answer for part (i)
Part 2:
The steps for construction are given below:
1. Step 1 is to draw an equilateral triangle of side 3 cm. This is shown in fig.32.21(a) below
2. Step 2 is to draw the circumcircle of this triangle. The method can be seen here. In this step, we get the position of the circumcentre 'O' also. This is shown in fig.b
3. Step 3 is to join all the three vertices to the centre O. These are the yellow lines in fig.c
Now draw lines perpendicular to the yellow lines through the vertices. These lines (shown in red colour in fig.c) will give the outer triangle
Solved example 32.9
Fig.32.23(a) shows the two tangents (red lines) drawn from an external point. And also the radii (green lines) through the points of contact.
(i) Prove that the tangents have the same length
(ii) In fig(b), the external point is joined to the center of the circle (cyan line). Prove that this cyan line bisects the angle between the radii
(iii) Prove that the cyan line bisects the angle between the tangents also
(iv) In fig(c), a chord is drawn between the points of contact (magenta line). prove that the cyan line is the perpendicular bisector of the magenta chord
Solution:
Part (i): The tangents in fig32.23 will be of equal length.We proved it earlier and wrote Theorem 32.2.
Part (ii): We have to prove that the white and grey colored angles in fig.32.24(b) below are equal
We have proved it already and wrote the first part of Theorem 32.3
Part (iii): We have to prove that the green and yellow colored angles in fig.32.24(b) above are equal
We have proved it already and wrote the second part of Theorem 32.3
Part (iv): We have to prove that the cyan line is perpendicular to the magenta chord.
♦ And also that it divides the magenta chord into two equal parts as shown in fig.32.24(c) above.
• We proved it in solved example 32.4. See fig.32.13. It is shown again below:
• We proved that TO is the perpendicular bisector of PQ
Solved example 32.10
(i) In the fig.32.25(a) below, the two magenta chords are perpendicular to each other.
Red colored tangents are drawn at the ends of those chords. Those red tangents meet at four points to give a quadrilateral. Prove that this quadrilateral is cyclic
(ii) What sort of quadrilateral will it be, if one chord is a diameter?
(iii) What sort of quadrilateral will it be, if both chords are diameters?
Solution:
Part (i):
1. Let the magenta chords be PQ and RS. See fig.32.25(b)
2. Red tangents are drawn at P, Q, R and S
• Those tangents meet at four points to give a quadrilateral ABCD
3. The tangent points P, Q, R and S are joined to the centre O by green radial lines
• So, if ∠POR = x, we will get: ∠PAR = (180-x)
• Similarly, if ∠SOQ = y, we will get: ∠SCQ = (180-y)
4. In the quadrilateral ABCD, ∠PAR and ∠SCQ are opposite angles
• Their sum is: [(180-x) + (180-y)] = [360-(x+y)]
5. Consider the arc RP. It is drawn from the ends of perpendicular chords
• Similarly, the arc QS is drawn from the ends of the same perpendicular chords
• If we align the two arcs together, they will form a semi-circle. We proved this as a solved example in a video presentation)
6. The central angles of the arcs are xo and yo. Since they form a semi -circle, we get: (x+y) = 180o
7. So the sum in (4) becomes: [360-(x+y)] = [360-180] = 180
8. That is., sum of opposite angles in the quadrilateral ABCD is 180o. So it is a cyclic quadrilateral. This is shown in fig.32.25(c)
Part (ii):
1. Let the magenta chords be PQ and RS. See fig.32.26(a)
• This time, RS is a diameter
2. Red tangents are drawn at P, Q, R and S
• Those tangents meet at four points to give a quadrilateral ABCD
♦ We have to find which type of quadrilateral is ABCD
■ Since PQ and RS are perpendicular chords, the quadrilateral ABCD will be cyclic. This we have proved in part (i)
3. Since RS is a diameter, the tangents at R and S will be parallel.
• That is., the opposite sides AD and BC are parallel
♦ This we proved in solved example 32.5 in the previous section
4. If the opposite are parallel in a quadrilateral, it will be a trapezium
■ If a trapezium is cyclic, it will be an isosceles trapezium.
The proof is given in this video presentation.
Another method:
1. The tangent points P, Q, R and S are joined to the centre O by green radial lines. See fig.32.26(b)
Note that, The radial lines for R and S will be OR and OS themselves. Because RS is a diameter.
• So, if ∠POR = x, we will get: ∠PAR = (180-x)
• Similarly, if ∠SOQ = y, we will get: ∠SCQ = (180-y)
2. Since RS is a diameter, the tangents at R and S will be parallel.
• That is., the opposite sides AD and BC are parallel
♦ This we proved in solved example 32.5 in the previous section
3. So we have two parallel lines cut by a transversal CD
• If the angle at C is (180-y), angle at D will be y (∵ AD and BC are parallel)
See proof in video presentation.
4. But from (6) in part (i), we have: (x+y) = 180
• So we get: y = (180-x)
5. But from (1) above, we have: angle at A = (180-x)
• So angles at A and D are equal
■ If AD and BC are parallel and angles at A and D are equal, it will be an isosceles trapezium.
Part (iii):
1. Let the magenta chords be PQ and RS. See fig.32.27(a)
• This time, both PQ and RS are diameters
2. So opposite sides AB and CD will be parallel
The opposite sides AD and BC will also be parallel
3. Since the opposite sides are parallel to the perpendicular lines PQ and RS, we get:
∠A = ∠B = ∠C = ∠D = 90o
• This is shown in fig.32.27(b)
4. Since the corner angles are all 90o, we get:
• Lengths AB = BC = CD = AD
5. From (2), (3) and (4), it is clear that, ABCD is a square.
In the next section, we will see more details about Tangents.
1. Consider the two tangents AP and BP in fig.32.15 below.
• OA and OB are radii.
♦ The tangents are drawn at the points of intersection of the radii with the circle
 |
| Fig.32.15 |
3. Now consider the quadrilateral OAPB. It is formed by the radii and the tangents
4. A and B are two opposite corners. The sum of the angles at those opposite corners = 90+90 = 180o
■ If the sum of opposite angles in any quadrilateral is 180o, that quadrilateral will be a cyclic quadrilateral,
• That is., all the four vertices of that quadrilateral will lie on a circle (We saw details here)
5. So we can draw a circle through all the four vertices of the quadrilateral OAPB. This circle is shown in green colour in fig.32.15(b) above
We can write the above result in the form of a theorem:
Theorem 32.4
• If a quadrilateral has the following four vertices:
♦ First vertex at the centre of a circle
♦ Second vertex at the exterior point from which two tangents are drawn to the circle
♦ Third vertex at one tangent point
♦ Fourth vertex at the other tangent point
■ Then that quadrilateral is cyclic
From the above, we get another useful result:
• In the fig.32.15(b) above, we considered the opposite vertices A and B. What about the other two opposite vertices O and P?
• Obviously, the sum of those two angles must also be 180o (∵ the sum of interior angles of any quadrilaterals is 360o)
This result is also useful for solving problems. We will write it as a theorem:
Theorem 32.5
1. In a circle, two radii are drawn
2. Tangents are drawn at the end point of each of those radii
• These tangents meet at P
3. Then sum of the following two angles is 180o
• Angle at O between the two radii
• Angle at P between the two tangents
Now we will see an application of this theorem in the form of a solved example
Solved example 32.6
Draw an equilateral triangle in such a way that it's all three sides are tangential to a circle of radius 3 cm
Solution:
A rough sketch is shown in fig.32.16(a) below:
 |
| Fig.32.16 |
• The triangle is to be equilateral. So all it's angles are 60
• Based on this rough fig., we must obtain details to make the actual construction
We will write the steps:
1. From the center O, draw radial lines OP, OQ and OR. This is shown in fig.b
2. Based on theorem 32.5, we get:
• ∠A + ∠ROP = 180 ⟹ 60 + ∠ROP = 180 ⟹ ∠ROP = 120
• ∠B + ∠POQ = 180 ⟹ 60 + ∠POQ = 180 ⟹ ∠POQ = 120
• ∠C + ∠ROQ = 180 ⟹ 60 + ∠ROQ = 180 ⟹ ∠ROQ = 120
• So the three central angles are all 120o. This is marked in fig.c
• We can now begin the actual construction
3. Step 1 is to draw a circle of radius 3 cm. Mark the center as 'O'. Then draw a radial line OP in any convenient direction. This is shown in fig.32.17(a) below:
 |
| Fig.32.17 |
• Draw OQ in such a way that it makes an angle of 120o with OP
• Draw OR in such a way that it makes an angle of 120o with OQ
• Then the third angle ∠POR will be naturally 120
This is shown in fig.32.17(b)
5. Step 3 is the final step. In this step we draw the tangents:
• Draw a line perpendicular to OP at P. This is the tangent at P
• Draw a line perpendicular to OQ at Q. This is the tangent at Q
• Draw a line perpendicular to OR at R. This is the tangent at R
■ The three tangents will meet at A, B and C, giving the required equilateral triangle. This is shown in fig.c
Draw a triangle of angles 40o, 60o and 80o in such a way that it's all three sides are tangential to a circle of radius 2.5 cm
Solution:
A rough sketch is shown in fig.32.18(a) below:
 |
| Fig.32.18 |
• The required angles are also marked in the fig.a
• Based on this rough fig., we must obtain details to make the actual construction
We will write the steps:
1. From the center O, draw radial lines OP, OQ and OR. This is shown in fig.b
2. Based on theorem 32.5, we get:
• ∠A + ∠ROP = 180 ⟹ 40 + ∠ROP = 180 ⟹ ∠ROP = 140
• ∠B + ∠POQ = 180 ⟹ 60 + ∠POQ = 180 ⟹ ∠POQ = 120
• ∠C + ∠ROQ = 180 ⟹ 80 + ∠ROQ = 180 ⟹ ∠ROQ = 100
• So the three central angles are 140o, 120o and 100o. This is marked in fig.c
• We can now begin the actual construction
3. Step 1 is to draw a circle of radius 2.5 cm. Mark the center as 'O'. Then draw a radial line OP in any convenient direction. This is shown in fig.32.19(a) below:
 |
| Fig.32.19 |
• Draw OQ in such a way that it makes an angle of 120o with OP
• Draw OR in such a way that it makes an angle of 100o with OQ
• Then the third angle ∠POR will be naturally 140o
This is shown in fig.32.19(b)
5. Step 3 is the final step. In this step we draw the tangents:
• Draw a line perpendicular to OP at P. This is the tangent at P
• Draw a line perpendicular to OQ at Q. This is the tangent at Q
• Draw a line perpendicular to OR at R. This is the tangent at R
■ The three tangents will meet at A, B and C, giving the required triangle. This is shown in fig.c
Solved example 32.8
In the fig.32.20(a), the blue triangle is equilateral. It's circumcircle is shown in green colour. All the sides of the red triangle are tangents to the green circle. The tangent points are the vertices of the blue triangle.
 |
| Fig.32.20 |
(ii) Draw the fig. with sides of the small triangle 3 cm
Solution:
Part 1:
Let ABC be the red triangle and PQR be the blue triangle. This is marked in fig.b
1. Mark the center O of the green circle. Draw radial lines OP, OQ and OR. They are shown in yellow colour
2. Given that the red lines are tangents at the vertices. So the yellow lines are perpendicular to the red lines. we can write:
AB is perpendicular to OP
BC is perpendicular to OQ
AC is perpendicular to OR
3. Each of the three blue lines PQ, QR and RP subtend an angle at the center O:
♦ PQ subtend ∠POQ
♦ QR subtend ∠QOR
♦ RP subtend ∠ROP
• But PQ = QR = RP (∵ they are sides of an equilateral triangle)
• So the angle subtended by them must be equal
♦ Thus we get: ∠POQ = ∠QOR = ∠ROP
• But the total angle at any point is 360o
♦ We can write: ∠POQ + ∠QOR + ∠ROP = 360o
• So, since the three angles are equal, we get: ∠POQ = ∠QOR = ∠ROP = 360⁄3 = 120o
4. Based on theorem 32.5, we can write:
• ∠A + ∠ROP = 180 ⟹ ∠A+ 120 = 180 ⟹ ∠A = 60
• ∠B + ∠POQ = 180 ⟹ ∠B+ 120 = 180 ⟹ ∠B = 60
• ∠C + ∠ROQ = 180 ⟹ ∠C+ 120 = 180 ⟹ ∠C = 60
• So the three angles of the red triangle are all 60o. This is marked in fig.b. Thus the red triangle ABC is an equilateral triangle.
5. Consider fig.c: AB and AC are two tangents from an exterior point A
(i) P and R are the tangent points. Then by theorem 32.2 that we saw in the previous section, AP = AR
(ii) In the same way, we have: BP = BQ and CQ = CR
(iii) Let: AP = AR = a
BP = BQ = b
CQ = CR = c
(iv) Since the triangle is equilateral, we have: AB = BC = CA ⟹ (a+b) = (b+c) = (c+a)
From (a+b) = (b+c), we get: a = c
From (b+c) = (c+a), we get: a = b
• So we have: a = b = c
6. P is the midpoint of AB (∵ a = b)
R is the midpoint of AC (∵ a = c)
• So the line PR cuts the sides AB and AC at their midpoints
• Then by theorem 18.5, (Division of triangles by parallel lines) we get:
♦ BC is parallel to PR
♦ BC is double the length of PR
In the same way:
♦ AC is parallel to PQ
♦ AC is double the length of PQ
AND
♦ AB is parallel to RQ
♦ AB is double the length of RQ
■ (4) and (6) together gives the answer for part (i)
Part 2:
The steps for construction are given below:
1. Step 1 is to draw an equilateral triangle of side 3 cm. This is shown in fig.32.21(a) below
 |
| Fig.32.22 |
3. Step 3 is to join all the three vertices to the centre O. These are the yellow lines in fig.c
Now draw lines perpendicular to the yellow lines through the vertices. These lines (shown in red colour in fig.c) will give the outer triangle
Solved example 32.9
Fig.32.23(a) shows the two tangents (red lines) drawn from an external point. And also the radii (green lines) through the points of contact.
 |
| Fig.32.23 |
(ii) In fig(b), the external point is joined to the center of the circle (cyan line). Prove that this cyan line bisects the angle between the radii
(iii) Prove that the cyan line bisects the angle between the tangents also
(iv) In fig(c), a chord is drawn between the points of contact (magenta line). prove that the cyan line is the perpendicular bisector of the magenta chord
Solution:
Part (i): The tangents in fig32.23 will be of equal length.We proved it earlier and wrote Theorem 32.2.
Part (ii): We have to prove that the white and grey colored angles in fig.32.24(b) below are equal
 |
| Fig.32.24 |
Part (iii): We have to prove that the green and yellow colored angles in fig.32.24(b) above are equal
We have proved it already and wrote the second part of Theorem 32.3
Part (iv): We have to prove that the cyan line is perpendicular to the magenta chord.
♦ And also that it divides the magenta chord into two equal parts as shown in fig.32.24(c) above.
• We proved it in solved example 32.4. See fig.32.13. It is shown again below:
 |
| Fig.32.13 |
Solved example 32.10
(i) In the fig.32.25(a) below, the two magenta chords are perpendicular to each other.
 |
| Fig.32.25 |
(ii) What sort of quadrilateral will it be, if one chord is a diameter?
(iii) What sort of quadrilateral will it be, if both chords are diameters?
Solution:
Part (i):
1. Let the magenta chords be PQ and RS. See fig.32.25(b)
2. Red tangents are drawn at P, Q, R and S
• Those tangents meet at four points to give a quadrilateral ABCD
3. The tangent points P, Q, R and S are joined to the centre O by green radial lines
• So, if ∠POR = x, we will get: ∠PAR = (180-x)
• Similarly, if ∠SOQ = y, we will get: ∠SCQ = (180-y)
4. In the quadrilateral ABCD, ∠PAR and ∠SCQ are opposite angles
• Their sum is: [(180-x) + (180-y)] = [360-(x+y)]
5. Consider the arc RP. It is drawn from the ends of perpendicular chords
• Similarly, the arc QS is drawn from the ends of the same perpendicular chords
• If we align the two arcs together, they will form a semi-circle. We proved this as a solved example in a video presentation)
6. The central angles of the arcs are xo and yo. Since they form a semi -circle, we get: (x+y) = 180o
7. So the sum in (4) becomes: [360-(x+y)] = [360-180] = 180
8. That is., sum of opposite angles in the quadrilateral ABCD is 180o. So it is a cyclic quadrilateral. This is shown in fig.32.25(c)
Part (ii):
1. Let the magenta chords be PQ and RS. See fig.32.26(a)
• This time, RS is a diameter
 |
| Fig.32.26 |
• Those tangents meet at four points to give a quadrilateral ABCD
♦ We have to find which type of quadrilateral is ABCD
■ Since PQ and RS are perpendicular chords, the quadrilateral ABCD will be cyclic. This we have proved in part (i)
3. Since RS is a diameter, the tangents at R and S will be parallel.
• That is., the opposite sides AD and BC are parallel
♦ This we proved in solved example 32.5 in the previous section
4. If the opposite are parallel in a quadrilateral, it will be a trapezium
■ If a trapezium is cyclic, it will be an isosceles trapezium.
The proof is given in this video presentation.
Another method:
1. The tangent points P, Q, R and S are joined to the centre O by green radial lines. See fig.32.26(b)
Note that, The radial lines for R and S will be OR and OS themselves. Because RS is a diameter.
• So, if ∠POR = x, we will get: ∠PAR = (180-x)
• Similarly, if ∠SOQ = y, we will get: ∠SCQ = (180-y)
2. Since RS is a diameter, the tangents at R and S will be parallel.
• That is., the opposite sides AD and BC are parallel
♦ This we proved in solved example 32.5 in the previous section
3. So we have two parallel lines cut by a transversal CD
• If the angle at C is (180-y), angle at D will be y (∵ AD and BC are parallel)
See proof in video presentation.
4. But from (6) in part (i), we have: (x+y) = 180
• So we get: y = (180-x)
5. But from (1) above, we have: angle at A = (180-x)
• So angles at A and D are equal
■ If AD and BC are parallel and angles at A and D are equal, it will be an isosceles trapezium.
Part (iii):
1. Let the magenta chords be PQ and RS. See fig.32.27(a)
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| Fig.32.27 |
2. So opposite sides AB and CD will be parallel
The opposite sides AD and BC will also be parallel
3. Since the opposite sides are parallel to the perpendicular lines PQ and RS, we get:
∠A = ∠B = ∠C = ∠D = 90o
• This is shown in fig.32.27(b)
4. Since the corner angles are all 90o, we get:
• Lengths AB = BC = CD = AD
5. From (2), (3) and (4), it is clear that, ABCD is a square.
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