Chapter 27.8 - Cyclic quadrilaterals - Solved examples

In the previous section we saw theorem 27.9 and it's converse. In this section we will see some solved examples.

Solved example 27.15
In the fig.27.46(a) below, A, B, C and D are four points on a circle. 

Fig.27.45

∠DBC = 55^o and ∠CAB = 45^o. Compute ∠BCD
Solution:

1. Consider arc BC in fig(b). It subtends ∠BAC (= 45^o) on the alternate arc. 

• The same arc subtends ∠BDC on the alternate arc. So ∠BDC = ∠BAC = 45^o. This is marked in fig(b)
2. Consider ΔBDC. We get ∠BCD = [180-(45+55)] =[180-100] = 80^o. (∵ sum of interior angles of a triangle is 180^o)
Thus we get the required angle. We can do a check by using theorem 27.9.   

3. Consider arc CD in fig(c). It subtends ∠CBD (= 55^o) on the alternate arc. 

• The same arc subtends ∠CAD on the alternate arc. So ∠CBD = ∠CAD = 55^o. This is marked in fig(c)
(i) Sum of opposite angles ∠BAD and ∠BCD = 45 + 55 + 80 = 180^o.          

Solved examples 27.16
In the fig.27.46(a) below, A, B, C and D are four points on a circle. 

Fig.27.46

AC and BD intersect at E in such a way that ∠BEC = 30^o. and ∠ECD = 20^o. Find ∠BAC
Solution:
1. ∠BEC and ∠BEA form a linear pair. So ∠BEA = 180 - ∠BEC = 180 -130 = 50^o.
2. ∠BEA and ∠CED are opposite angles, and are hence equal. So we get ∠BEA = ∠CED = 50^o.

3. Consider ΔCED. We get ∠EDC = [180-(50+20)] =[180-70] = 110^o. (∵ sum of interior angles of a triangle is 180^o)

4. Consider the major arc BC in fig(c). It subtends ∠BDC (= 110^o) on the alternate arc. 

• The same arc subtends ∠BAC on the alternate arc. So ∠BAC = ∠BDC = 110^o. This is marked in fig(c). Thus we get the required angle.

Solved example 27.17
In the fig.27.47(a) below, ABCD is a square. 

Fig.27.47

Determine ∠DPC
Solution: 
1. Draw the diagonal AC of the square. 
2. A diagonal of a square will bisect the angles at the corners. So we get:
∠DAC = ∠BAC = 45^o.
3. Consider the quadrilateral ACPD. It is a cyclic quadrilateral. The sum of opposite angles = 180^o.
4. So we get: ∠DPC + ∠DAC = 180^o ⇒ ∠DPC + 45 = 180 ⇒ ∠DPC = 180 - 45 = 135^o.

---

Now we will see an important result related to cyclic quadrilaterals. We will learn it in steps: 

1. Fig.27.48(a) below shows a cyclic quadrilateral ABCD. 

Fig.27.48

2. The side AB is extended along towards the right up to point E. So ∠CBE (shown in red colour) becomes an exterior angle of the cyclic quadrilateral. We can write:
■ ∠CBE is the exterior angle of the cyclic quadrilateral ABCD at the vertex B. 
3. For the vertex B, the opposite vertex is D
• So, for the vertex B, ∠ADC (shown in yellow colour) is the 'interior angle at the opposite vertex'
4. Thus we have three quantities:
(i) A vertex B  (ii) Exterior angle at that vertex B  (iii) Interior angle at D, which is the opposite vertex of B  
• We want to know the relation between (ii) and (iii)
5. Consider the interior angle at B. It is shown in blue colour in fig (b)
• Blue + Red will obviously be 180^o (∵ they form a linear pair)
• So we can write: ∠CBE + ∠ABC = 180^o
6. Yellow and Blue are opposite angles of a cyclic quadrilateral. So their sum will be 180^o. 
• We can write: ∠ADC + ∠ABC = 180^o
7. From (5) we get: ∠ABC = 180 – ∠CBE
• Substituting this in (6) we get:
∠ADC + (180 – ∠CBE) = 180
⇒ ∠ADC – ∠CBE = 180 – 180  
⇒ ∠ADC – ∠CBE = 0 
⇒ ∠ADC = ∠CBE
8. So we can write: 
• The exterior angle at B is equal to the interior angle at opposite vertex
9. Now consider fig (c). The side CB is extended upto F
• ∠ABF is an exterior angle at vertex B. So is ∠CBE
• But we can see that the two are equal because, they are opposite angles. 
• So, at a vertex, there will be only one value for an exterior angle
10. We can write the above results in a general form:
• Consider any vertex of a cyclic quadrilateral. 
• There will be an exterior angle at that vertex
• That exterior angle will be equal to the interior angle at the opposite vertex

---

Some solved examples on cyclic quadrilaterals are shown in the form of a video presentation at the following links:
Trapezium Cyclic or not

Non-rectangular parallelogram Cyclic or not

Video presentation 1 

Video presentation 2 

Video presentation 3 

Video presentation 4 

Parallelogram with one vertex at center

---

In the next section, we will learn about Multiplication of Chords.

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Maths lessons. blogspot.in - All Rights Reserved

Chapter 27.7 - Cyclic Quadrilaterals

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.

Fig.27.42

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 = 180^o.
5. In a similar way, by drawing the diagonal BD as in fig(c), we will get:
∠BAD + ∠BCD = 180^o

---

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

■ We will now analyse each of them. 
1. Consider fig.27.44(a) below. 

Fig.27.44

• 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 = 180^o.
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 180^o. 
That is.,  ∠ABC + ∠ADE < 180^o.
■ 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 180^o.

---

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 = 180^o.
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 180^o. 
That is.,  ∠ABC + ∠ADC > 180^o.
■ 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 180^o.

---

Now we can apply the above two findings to the practical situation:

1. We are given a quadrilateral ABCD with all the interior angles
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 180^o
■ 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 180^o, one vertex will fall outside the circle
• If any of the sum is greater than 180^o, one vertex will fall inside the circle
• So if it is exact 180^o, 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.

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Maths lessons. blogspot.in - All Rights Reserved