In the previous section we saw the details about Segments and their angles. In this section we will learn about Cyclic quadrilaterals.
1. In the fig.27.42(a) below, ABCD is a quadrilateral. All it's four vertices are on a circle.
• We want to know whether there is any relation between the angles at the vertices.
2. For that, first draw the diagonal AC as shown in fig(b). The diagonal is a chord. It separates the circle into two segments. Segment ABC and segment ADC
3. The ∠ABC is the unique angle of segment ABC
The ∠ADC is the unique angle of segment ADC
4. Each of the segments is the alternate segment of the other.
• So we get ∠ABC + ∠ADC = 180o.
5. In a similar way, by drawing the diagonal BD as in fig(c), we will get:
∠BAD + ∠BCD = 180o
If all the vertices of a quadrilateral are on a circle, then the opposite angles of that quadrilateral are supplementary.
1. Consider any quadrilateral. It has four vertices.
2. Take out any three of them. We surely can draw a circle through those three.
• Because we have learned how to draw a circle through any three points, if they are not collinear. Details here.
3. Now, after drawing the circle through the three vertices, we may find ourselves in any one of the two situations below:
• The fourth vertex is outside the circle. An example is shown in fig.27.43(a) below
• The fourth vertex is inside the circle. An example is shown in fig.27.43(b) below
■ We will now analyse each of them.
1. Consider fig.27.44(a) below.
• It is the same 27.43(a) that we saw above.
• The only modification is this:
The point where CD cuts the circle is named as E. And A is joined to E by a red line
2. Now the vertices A, B, C and E are on the circle. So we can write:
∠ABC + ∠AEC = 180o.
3. Consider ΔAED. The ∠AEC is an exterior angle of this ΔAED
• Exterior angle = sum of remote interior angles
So ∠AEC = ∠DAE + ∠ADE
4. Consider the right side of the above equation.
• Only when we add a quantity of '∠DAE' to ∠ADE, the right side becomes equal to ∠AEC
• So ADE is always less than AEC. That is., ∠ADE < ∠AEC
5. Now consider the left side of (2). If we put ∠ADE instead of ∠AEC , the sum on the left side will always be less than 180o.
That is., ∠ABC + ∠ADE < 180o.
■ Let us write a summary of the above discussion:
• We are given a quadrilateral ABCD
• We take out any three of it's vertices. Let them be A, B and C
• It is possible to draw a circle through any three points. So we draw a circle through A, B and C
• We find that D is outside the circle
• Then we can write this:
The sum of the angles at the 'outside vertex D' and 'it's opposite vertex B' will always be less than 180o.
The only modification is this:
CD is extended to meet the circle at E. And A is joined to E
2. Now the vertices A, B, C and E are on the circle. So we can write:
∠ABC + ∠AED = 180o.
3. Consider ΔAED. The ∠ADC is an exterior angle of this ΔAED
• Exterior angle = sum of remote interior angles
So ∠ADC = ∠DAE + ∠AED
⇒ ∠AED = ∠ADC - ∠DAE
4. Consider the right side of the above equation.
• Only when we subtract a quantity of '∠DAE' from ∠ADC, the right side becomes equal to ∠AED
• So ∠AED is always less than ∠ADC. That is., ∠ADC > ∠AED
5. Now consider the left side of (2). If we put ∠ADC instead of ∠AED , the sum on the left side will always be greater than 180o.
That is., ∠ABC + ∠ADC > 180o.
■ Let us write a summary of the above discussion:
• We are given a quadrilateral ABCD
• We take out any three of it's vertices. Let them be A, B and C
• It is possible to draw a circle through any three points. So we draw a circle through A, B and C
• We find that D is inside the circle
• Then we can write this:
The sum of the angles at the 'inside vertex D' and 'it's opposite vertex B' will always be greater than 180o.
2. We take out the two pairs of opposite vertices:
• First pair is A and C
• Second pair is B and D
3. We find the sum of angles of each pair. That is:
• ∠A + ∠C
• ∠B + ∠D
4. We find that both the sums are exactly equal to 180o
■ In such a situation, do all the four vertices fall on a circle?
Ans: Indeed they do. Let us see the reason:
• If any of the sum is less than 180o, one vertex will fall outside the circle
• If any of the sum is greater than 180o, one vertex will fall inside the circle
• So if it is exact 180o, it is neither greater than nor less than 180. That means, all the vertices lie on the circle
■ So we got the proof for the converse of theorem 27.9. Let us write it:
1. In the fig.27.42(a) below, ABCD is a quadrilateral. All it's four vertices are on a circle.
• We want to know whether there is any relation between the angles at the vertices.
 |
| Fig.27.42 |
3. The ∠ABC is the unique angle of segment ABC
The ∠ADC is the unique angle of segment ADC
4. Each of the segments is the alternate segment of the other.
• So we get ∠ABC + ∠ADC = 180o.
5. In a similar way, by drawing the diagonal BD as in fig(c), we will get:
∠BAD + ∠BCD = 180o
So we can write the above result as a theorem
Theorem 27.9:If all the vertices of a quadrilateral are on a circle, then the opposite angles of that quadrilateral are supplementary.
■ Now we want to know whether the converse of the theorem is true. That is., if opposite angles of a quadrilateral are supplementary, will all the four vertices lie on a circle?
Let us check:1. Consider any quadrilateral. It has four vertices.
2. Take out any three of them. We surely can draw a circle through those three.
• Because we have learned how to draw a circle through any three points, if they are not collinear. Details here.
3. Now, after drawing the circle through the three vertices, we may find ourselves in any one of the two situations below:
• The fourth vertex is outside the circle. An example is shown in fig.27.43(a) below
• The fourth vertex is inside the circle. An example is shown in fig.27.43(b) below
 |
| Fig.27.43 |
1. Consider fig.27.44(a) below.
 |
| Fig.27.44 |
• The only modification is this:
The point where CD cuts the circle is named as E. And A is joined to E by a red line
2. Now the vertices A, B, C and E are on the circle. So we can write:
∠ABC + ∠AEC = 180o.
3. Consider ΔAED. The ∠AEC is an exterior angle of this ΔAED
• Exterior angle = sum of remote interior angles
So ∠AEC = ∠DAE + ∠ADE
4. Consider the right side of the above equation.
• Only when we add a quantity of '∠DAE' to ∠ADE, the right side becomes equal to ∠AEC
• So ADE is always less than AEC. That is., ∠ADE < ∠AEC
5. Now consider the left side of (2). If we put ∠ADE instead of ∠AEC , the sum on the left side will always be less than 180o.
That is., ∠ABC + ∠ADE < 180o.
■ Let us write a summary of the above discussion:
• We are given a quadrilateral ABCD
• We take out any three of it's vertices. Let them be A, B and C
• It is possible to draw a circle through any three points. So we draw a circle through A, B and C
• We find that D is outside the circle
• Then we can write this:
The sum of the angles at the 'outside vertex D' and 'it's opposite vertex B' will always be less than 180o.
Now we consider the other case. That is., D lies inside the circle
1. This is shown in fig.27.44(b). it is the same fig.27.43(b) that we saw earlier.The only modification is this:
CD is extended to meet the circle at E. And A is joined to E
2. Now the vertices A, B, C and E are on the circle. So we can write:
∠ABC + ∠AED = 180o.
3. Consider ΔAED. The ∠ADC is an exterior angle of this ΔAED
• Exterior angle = sum of remote interior angles
So ∠ADC = ∠DAE + ∠AED
⇒ ∠AED = ∠ADC - ∠DAE
4. Consider the right side of the above equation.
• Only when we subtract a quantity of '∠DAE' from ∠ADC, the right side becomes equal to ∠AED
• So ∠AED is always less than ∠ADC. That is., ∠ADC > ∠AED
5. Now consider the left side of (2). If we put ∠ADC instead of ∠AED , the sum on the left side will always be greater than 180o.
That is., ∠ABC + ∠ADC > 180o.
■ Let us write a summary of the above discussion:
• We are given a quadrilateral ABCD
• We take out any three of it's vertices. Let them be A, B and C
• It is possible to draw a circle through any three points. So we draw a circle through A, B and C
• We find that D is inside the circle
• Then we can write this:
The sum of the angles at the 'inside vertex D' and 'it's opposite vertex B' will always be greater than 180o.
Now we can apply the above two findings to the practical situation:
1. We are given a quadrilateral ABCD with all the interior angles2. We take out the two pairs of opposite vertices:
• First pair is A and C
• Second pair is B and D
3. We find the sum of angles of each pair. That is:
• ∠A + ∠C
• ∠B + ∠D
4. We find that both the sums are exactly equal to 180o
■ In such a situation, do all the four vertices fall on a circle?
Ans: Indeed they do. Let us see the reason:
• If any of the sum is less than 180o, one vertex will fall outside the circle
• If any of the sum is greater than 180o, one vertex will fall inside the circle
• So if it is exact 180o, it is neither greater than nor less than 180. That means, all the vertices lie on the circle
■ So we got the proof for the converse of theorem 27.9. Let us write it:
Converse of theorem 27.9:
If opposite angles of a quadrilateral are supplementary, then all the four vertices of that quadrilateral will lie on a circle
In the next section, we will see some solved examples.
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