In the previous section we saw the division of a triangle by the line joining a vertex to any point on the opposite side. We saw some solved examples also. In this section, we will see a more advanced case.
• Consider the ΔABC in fig.14.28(a).
• It has sides AC = x1 and BC = y1
• CD is the bisector of angle ACB. So we have ∠ACD = ∠BCD = half of ∠ACB.
• As those angles are equal, we will denote them as α. This is shown in the fig.b.
• The fig.b shows many more details:
• The angle bisector CD, splits the ΔABC into two: ΔACD and ΔBCD.
• Perpendicular DF is drawn from F to AC. Similarly, perpendicular DE is drawn from D to BC.
• When the perpendicular lines DF and DE are drawn, we get two new triangles: ΔCDF and ΔCDE. Let us analyse these two triangles:
• In any triangle, the sum of interior angles = 180o
• So in ΔCDF, 90 + α + ∠CDF = 180. ⇒ ∠CDF = 180 – (90 + α) = 90 – α
• Similarly, in ΔCDE, 90 + α + ∠CDE = 180. ⇒ ∠CDE = 180 – (90 + α) = 90 – α
• So we find that ∠CDF = ∠CDE = 90 – α
• Since both are equal, we will denote them as β
• So, in ΔCDF, we have two angles α and β, and an included side CD. Those same angles and included side are present in ΔCDE also. So this is a case of ASA congruence.
• Let us write the correspondence:
• E↔F since both are 90o. C↔C since both are α. D↔D since both are β
• So the correspondence is FCD↔ECD
• The corresponding sides are FC↔EC, CD↔CD and FD↔ED
• So sides FD and ED are equal in length. We will denote them as 'h'
• Now consider the original two triangles: ΔADC and ΔBDC. We can derive a relation between their area:
We will write the above findings in the form of a theorem. Usually, theorems are written in a single sentence. Here we will write step by step.
Theorem 14.7
1 The angle bisector is drawn at a vertex
2 The lengths of the sides meeting at that vertex are in the ratio x1:y1
3 The angle bisector divides the triangle into two smaller triangles
4 The areas of these two smaller triangles will also be in the ratio x1:y1
5 While taking ratio of areas, x1 and y1 should be assigned to the corresponding triangles
The converse of this theorem can also be written:
Theorem 14.8
1 The angle bisector is drawn at a vertex
2 This angle bisector divides the triangle into two smaller triangles
3 The areas of these two smaller triangles are in the ratio x1:y1
4 The lengths of the sides meeting at that vertex will also be in the ratio x1:y1
5 While taking ratio of areas, x1 and y1 should be assigned to the corresponding triangles
The above two theorems is a good example to demonstrate a Theorem and it's converse.
Note that it is a sort of 'reversal'.
• In Theorem 14.7, if sides are in the ratio x1: y1, then the areas are also in the ratio x1: y1
• In Theorem 14.8, if areas are in the ratio x1: y1, then the sides are also in the ratio x1: y1
Now we will see a very interesting relation between the above theorem 14.7 and the previous theorem 14.6, which we derived based on fig.14.26(a). We will draw the two triangles together:
• In fig(a), CD is any line drawn from the vertex C to BD. Ratio of the areas of the two triangles ADC and BDC = x : y. [x1 and y1 has no role to play here]
• If this line CD is not 'just any line', but the bisector of the angle at C, then the ratio of the areas will be in he ratio x1 : y1
• But which ever line we draw from the vertex, the ratio of the areas is in the ratio x : y
• So when the line CD is the bisector of the angle at C, we have two equalities:
1 ar (ADC): ar (BDC) = x1 : y1
2 ar (ADC) : ar (BDC) = x : y
• So, when CD is the bisector of the angle, we have: x1 : y1 = x : y
• We can write this in the form of a theorem:
Theorem 14.9
1 The angle bisector is drawn at a vertex
2 The lengths of the sides meeting at that vertex are in the ratio x1 : y1
3 The angle bisector splits the opposite side into two line segments
4 The ratio of the lengths of these two line segments will also be equal to x1 : y1
The converse of this theorem can also be written:
Theorem 14.10
1 A line is drawn from a vertex to the opposite side
2 The lengths of the sides meeting at that vertex are in the ratio x1 : y1
3 The line splits the opposite side into two line segments
4 The ratio of the lengths of these two line segments are also equal to x1 : y1
5 Then the line drawn is not 'just any line', it is the bisector of the angle at that vertex
Now we will see a practical application of the above theorem. In the fig.14.30, AB is a line 8 cm long. We want to divide it in the ratio 5 : 4.
For that, we form a triangle with AB as the base. The other two sides must be 5 cm and 4 cm in length. In the fig.14.30(b), AC = 5 cm and BC = 4 cm. Consider the angle bisector CE drawn at the vertex C (fig.c). According to the Theorem 14.9, the point E will divide AB in the ratio 5:4. So we have an easy method to divide a line in any given ratio.
Now we will discuss how to draw the angle bisector CE.
• We will start with fig(b). Extend AC to any convenient point D'.
• With C as center, and CB as radius, draw an arc which cuts CD' at D.
• Then CD will be equal to CB. This is shown in fig(d).
• Now join B and D. Through C, draw CE, parallel to DB.
■ CE will be the angle bisector of ∠ACB.
Proof:
• BCD is an isosceles triangle since CB = CD
• Let us denote the two base angles as α
• ∠ECB = α [∵ ∠ECB and ∠CBD are alternate interior angles. Two parallel lines CE and DB cut by a transversal BC]
• ∠ACE = α [∵ ∠ACE and ∠CDB are corresponding angles. Two parallel lines CE and DB cut by a transversal BC]
• So we get ∠ACE = ∠BCE. That means, CE is the angle bisector of ∠ACB
What if we want to split the 8 cm long line in the ratio 3:4?
In this case also we can follow the above procedure. We form a triangle with AB as base. The two sides should be 3 cm and 4 cm. But we will not be able to form a triangle with the three sides: 8 cm, 3 cm and 4 cm. This is because 8 > (3 + 4). Details can be seen here.
So we take a convenient equivalent ratio of 3:4. We know that 6: 8 is an equivalent ratio of 3:4. So our triangle will have base = 8 cm, and the other two sides 6 cm and 8 cm. The fig.14.31 below shows the required construction.
• Extend AC to any convenient point D'
• With C as center, and CB as radius, draw an arc which cuts CD' at D.
• Then CD will be equal to CB
• Now join B and D. Through C, draw CE, parallel to DB
■ CE will be the angle bisector of ∠ACB
Now we will see some solved examples:
Solved example 14.19
Draw a line 7 cm long, and divide it in the ratio 5:4
Solution:
The required construction is shown in the fig.14.32 below:
The steps used in the construction are:
• Construct triangle ABC with sides AB = 7 cm, AC = 5 cm, and BC = 4 cm
• Extend AC to any convenient point D'
• With C as center, and CB as radius, draw an arc which cuts CD' at D.
• Then CD will be equal to CB
• Now join B and D. Through C, draw CE, parallel to DB
■ CE will be the angle bisector of ∠ACB
■ AE will be equal to 5/9 of 7. And BE will be equal to 4/9 of 7. That is., AE : BE = 5:4
In the next section we will see more solved examples.
 |
| Fig.14.28 |
• It has sides AC = x1 and BC = y1
• CD is the bisector of angle ACB. So we have ∠ACD = ∠BCD = half of ∠ACB.
• As those angles are equal, we will denote them as α. This is shown in the fig.b.
• The fig.b shows many more details:
• The angle bisector CD, splits the ΔABC into two: ΔACD and ΔBCD.
• Perpendicular DF is drawn from F to AC. Similarly, perpendicular DE is drawn from D to BC.
• When the perpendicular lines DF and DE are drawn, we get two new triangles: ΔCDF and ΔCDE. Let us analyse these two triangles:
• In any triangle, the sum of interior angles = 180o
• So in ΔCDF, 90 + α + ∠CDF = 180. ⇒ ∠CDF = 180 – (90 + α) = 90 – α
• Similarly, in ΔCDE, 90 + α + ∠CDE = 180. ⇒ ∠CDE = 180 – (90 + α) = 90 – α
• So we find that ∠CDF = ∠CDE = 90 – α
• Since both are equal, we will denote them as β
• So, in ΔCDF, we have two angles α and β, and an included side CD. Those same angles and included side are present in ΔCDE also. So this is a case of ASA congruence.
• Let us write the correspondence:
• E↔F since both are 90o. C↔C since both are α. D↔D since both are β
• So the correspondence is FCD↔ECD
• The corresponding sides are FC↔EC, CD↔CD and FD↔ED
• So sides FD and ED are equal in length. We will denote them as 'h'
• Now consider the original two triangles: ΔADC and ΔBDC. We can derive a relation between their area:
We will write the above findings in the form of a theorem. Usually, theorems are written in a single sentence. Here we will write step by step.
Theorem 14.7
1 The angle bisector is drawn at a vertex
2 The lengths of the sides meeting at that vertex are in the ratio x1:y1
3 The angle bisector divides the triangle into two smaller triangles
4 The areas of these two smaller triangles will also be in the ratio x1:y1
5 While taking ratio of areas, x1 and y1 should be assigned to the corresponding triangles
The converse of this theorem can also be written:
Theorem 14.8
1 The angle bisector is drawn at a vertex
2 This angle bisector divides the triangle into two smaller triangles
3 The areas of these two smaller triangles are in the ratio x1:y1
4 The lengths of the sides meeting at that vertex will also be in the ratio x1:y1
5 While taking ratio of areas, x1 and y1 should be assigned to the corresponding triangles
The above two theorems is a good example to demonstrate a Theorem and it's converse.
Note that it is a sort of 'reversal'.
• In Theorem 14.7, if sides are in the ratio x1: y1, then the areas are also in the ratio x1: y1
• In Theorem 14.8, if areas are in the ratio x1: y1, then the sides are also in the ratio x1: y1
Now we will see a very interesting relation between the above theorem 14.7 and the previous theorem 14.6, which we derived based on fig.14.26(a). We will draw the two triangles together:
 |
| Fig.14.29 |
• If this line CD is not 'just any line', but the bisector of the angle at C, then the ratio of the areas will be in he ratio x1 : y1
• But which ever line we draw from the vertex, the ratio of the areas is in the ratio x : y
• So when the line CD is the bisector of the angle at C, we have two equalities:
1 ar (ADC): ar (BDC) = x1 : y1
2 ar (ADC) : ar (BDC) = x : y
• So, when CD is the bisector of the angle, we have: x1 : y1 = x : y
• We can write this in the form of a theorem:
Theorem 14.9
1 The angle bisector is drawn at a vertex
2 The lengths of the sides meeting at that vertex are in the ratio x1 : y1
3 The angle bisector splits the opposite side into two line segments
4 The ratio of the lengths of these two line segments will also be equal to x1 : y1
The converse of this theorem can also be written:
Theorem 14.10
1 A line is drawn from a vertex to the opposite side
2 The lengths of the sides meeting at that vertex are in the ratio x1 : y1
3 The line splits the opposite side into two line segments
4 The ratio of the lengths of these two line segments are also equal to x1 : y1
5 Then the line drawn is not 'just any line', it is the bisector of the angle at that vertex
Now we will see a practical application of the above theorem. In the fig.14.30, AB is a line 8 cm long. We want to divide it in the ratio 5 : 4.
 |
| Fig.14.30 |
Now we will discuss how to draw the angle bisector CE.
• We will start with fig(b). Extend AC to any convenient point D'.
• With C as center, and CB as radius, draw an arc which cuts CD' at D.
• Then CD will be equal to CB. This is shown in fig(d).
• Now join B and D. Through C, draw CE, parallel to DB.
■ CE will be the angle bisector of ∠ACB.
Proof:
• BCD is an isosceles triangle since CB = CD
• Let us denote the two base angles as α
• ∠ECB = α [∵ ∠ECB and ∠CBD are alternate interior angles. Two parallel lines CE and DB cut by a transversal BC]
• ∠ACE = α [∵ ∠ACE and ∠CDB are corresponding angles. Two parallel lines CE and DB cut by a transversal BC]
• So we get ∠ACE = ∠BCE. That means, CE is the angle bisector of ∠ACB
What if we want to split the 8 cm long line in the ratio 3:4?
In this case also we can follow the above procedure. We form a triangle with AB as base. The two sides should be 3 cm and 4 cm. But we will not be able to form a triangle with the three sides: 8 cm, 3 cm and 4 cm. This is because 8 > (3 + 4). Details can be seen here.
So we take a convenient equivalent ratio of 3:4. We know that 6: 8 is an equivalent ratio of 3:4. So our triangle will have base = 8 cm, and the other two sides 6 cm and 8 cm. The fig.14.31 below shows the required construction.
 |
| Fig.14.31 |
• With C as center, and CB as radius, draw an arc which cuts CD' at D.
• Then CD will be equal to CB
• Now join B and D. Through C, draw CE, parallel to DB
■ CE will be the angle bisector of ∠ACB
Now we will see some solved examples:
Solved example 14.19
Draw a line 7 cm long, and divide it in the ratio 5:4
Solution:
The required construction is shown in the fig.14.32 below:
 |
| Fig.14.32 |
• Construct triangle ABC with sides AB = 7 cm, AC = 5 cm, and BC = 4 cm
• Extend AC to any convenient point D'
• With C as center, and CB as radius, draw an arc which cuts CD' at D.
• Then CD will be equal to CB
• Now join B and D. Through C, draw CE, parallel to DB
■ CE will be the angle bisector of ∠ACB
■ AE will be equal to 5/9 of 7. And BE will be equal to 4/9 of 7. That is., AE : BE = 5:4
In the next section we will see more solved examples.
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