CHAPTER 7 TRIANGLES
NCERT Class 9 Mathematics Textbook - 2018- for Blind Students made Screen Readable by Professor T K Bansal.
Remarks by DR T K Bansal:
(i) This chapter contains 60 figures. Do not be scared; each and every figure is fully described. We are sure that if you spend a little time on these diagrams and try to make these figures on your tactile geometry board, you will really enjoy the chapter.
(ii) Usually, the captions of the figures are written below the figure, however, I feel that for a visually impaired student it is better if the captions are given just above the figures. It is for this reason, the captions are given above the figures.
Table of symbols used in this chapter:
| Symbol | Known As | Used for | Description | Word Shortcut | <-> Corresponds to A both sided arrow ≅ Congruent to 3 parallel horizontal lines CPCT Corresponding parts of congruent triangles Is an abbreviation |
7.1 Introduction
NCERT Class 9 Mathematics Textbook for Blind Students made Screen Readable by Professor T K Bansal.
We have studied about triangles and their various properties in our earlier classes. We know that a closed figure formed by three intersecting straight lines is called a triangle. (‘Tri’ means ‘three’). A triangle has three sides, three angles and three vertices. For example, in Triangle ABC, denoted by ΔABC (see Figure 7.1).
Figure 7.1
Line segments AB, BC and CA are the three sides; ∠A, ∠B and ∠C are the three angles; and Points A, B and C are the three vertices. In Chapter 6 also, we have studied some properties of triangles. In this chapter, we will study in detail about the congruence of triangles, rules of congruence, some more properties of triangles and inequalities in a triangle. We have already verified most of these properties in earlier classes. We will now prove some of them.
7.2 Congruence of triangles
NCERT Class 9 Mathematics Textbook for Blind Students made Screen Readable by Professor T K Bansal.
We have observed that two copies of our photographs of the same size are identical. Similarly, two bangles of the same size, two ATM cards issued by the same bank are identical. We may recall that on placing a one-rupee coin on another minted in the same year, cover each other completely.
I am sure that we remember what such figures are called. Indeed, they are called congruent figures (‘congruent’ means equal in all respects, or figures whose shapes and sizes are both the same).
Let us now, draw two circles of the same radius and place one on the other. What do we observe? They cover each other completely and we call them congruent circles.
Let us repeat this activity by placing one square on the other with sides of the same measure (see Figure 7.2) or by placing two equilateral triangles of equal sides on each other. We will observe that the squares are congruent to each other and so are the equilateral triangles.
Figure 7.2
Wee may wonder why we are studying congruence! We all must have seen the ice tray in our refrigerator. We observe that the molds for making ice are all congruent. The cast used for molding in the tray also has congruent depressions (maybe all are rectangular or all circular or all triangular). So, whenever identical objects have to be produced, the concept of congruence is used in making the cast.
Sometimes, we may find it difficult to replace the refill in our pen with a new one and this is so when the new refill is not of the same size as the one, we want to remove. Obviously, if the two refills are identical or congruent, the new refill fits well in our pen.
So, we can find numerous examples where congruence of objects is applied in daily life situations.
Can you think of some more examples of congruent figures?
Now, which of the following figures are not congruent to the square in Fig 7.3 (i) :
Fig 7.3
The large squares in Figure 7.3 (ii) and (iii) are obviously not congruent to the one in Fig 7.3 (i), but the square in Fig 7.3 (iv) is congruent to the one given in Fig 7.3 (i).
Let us now discuss the congruence of two triangles.
We already know that two triangles are congruent if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle.
Now, which of the triangles given below are congruent to ΔABC in Figure 7.4 (i)?
Fig 7.4 (i)
Fig 7.4 (ii)
Fig 7.4 (iii)
Fig 7.4 (iv)
Fig 7.4 (v)
Let us cut out each of these triangles from Figure 7.4 (ii) to (v) and turn them around and try to cover ΔABC. We observe that triangles in Figure 7.4 (ii), (iii) and (iv) are congruent to ΔABC while ΔTSU of Fig 7.4 (v) is not congruent to ΔABC.
If ΔPQR is congruent to ΔABC, we write ΔPQR ≅ ΔABC.
Notice that when ΔPQR ≅ ΔABC, then sides of ΔPQR fall on corresponding equal sides of ΔABC and so is the case for the angles.
That is, side PQ covers side AB, side QR covers side BC and side RP covers side CA; and
∠P covers ∠A, ∠Q covers ∠B and ∠R covers ∠C.
Also, there is a one-one correspondence between the vertices. That is, P corresponds to A, Q to B, R to C and so on which is written as
P ↔ A; Q ↔ B; R ↔ C; ( the symbol ↔’ stands for ‘corresponds to’)
Note that under this correspondence, ΔPQR ≅ ΔABC; but it will not be correct to write ΔQRP ≅ ΔABC. ( the symbol ‘≅’ is like three small horizontal parallel lines, please note that the screen reader is reading it as ‘approximately equal to’, we must read it as ‘congruent to’ )
Similarly, for Figure 7.4 (iii),
FD ↔ AB; dE ↔ BC and eF ↔ cA,
and
vertex F ↔ vertex A; vertex D ↔ vertex B; and vertex E ↔ vertex C
So, ΔFDE ≅ ΔABC but writing ΔDEF ≅ ΔABC is not correct.
Give the correspondence between the triangle in Figure 7.4 (iv) and ΔABC.
So, it is necessary to write the correspondence of vertices correctly for writing of congruence of triangles in symbolic form.
Note that in congruent triangles corresponding parts are equal and we write in short ‘CPCT’ for corresponding parts of congruent triangles.
7.3 Criteria for Congruence of triangles
NCERT Class 9 Mathematics Textbook for Blind Students made Screen Readable by Professor T K Bansal.
In our earlier classes, we have learnt four different criteria for congruence of triangles. Let us recall them.
Draw two triangles with one side 3 cm. Are these triangles congruent? Observe that they are not congruent (see Figure 7.5(i) and Fig 7.5(ii)).
Figure 7.5(i)
Figure 7.5(ii)
Now, draw two triangles with one side 4 cm and one angle 50° (see Figure 7.6). Are they congruent?
Fig 7.6
See that these two triangles are not congruent.
Repeat this activity with some more pairs of triangles.
So, equality of one pair of sides or one pair of sides and one pair of angles is not sufficient to give us congruent triangles.
What would happen if the other pair of arms (sides) of the equal angles were also equal?
In Fig 7.7, Side BC = Side QR, ∠B = ∠Q and also, Side AB = Side PQ. Now, what can you say about congruence of ΔABC and ΔPQR?
Fig 7.7
Recall from your earlier classes that, in this case, the two triangles are congruent.
Verify this for ΔABC and ΔPQR in Figure 7.7.
Repeat this activity with other pairs of triangles. Do you observe that the equality of both sides and the included angle is just enough for the congruence of triangles? Yes, it is enough.
This is the first criterion for congruence of triangles.
Axiom 7.1 (SAS congruence rule)
Two triangles are congruent if both sides and the included angle of one triangle, are equal to the both sides and the included angle of the other triangle.
This result cannot be proved with the help of previously known results and so it is accepted true as an axiom (see Appendix 1).
Let us now take some examples.
Example 1
In Figure 7.8, Side OA = Side OB and Side OD = Side OC.
Figure 7.8
Show that
(i)ΔAOD ≅ ΔBOC and
(ii) Side AD ∥ BC.
Solution :
(i) We can observe that in ΔAOD and ΔBOC,
Side OA = Side OB (Given)
Side OD = Side OC (Given)
Also, since ∠AOD and ∠BOC form a pair of vertically opposite angles, we have
∠AOD = ∠BOC.
So, ΔAOD ≅ΔBOC (by the SAS congruence rule)
(ii) In congruent triangles AOD and BOC, the other corresponding parts are also equal.
So, ∠OAD = ∠OBC and these form a pair of alternate angles for line segments AD and BC.
Therefore, Side AD ∥ Side BC. ( symbol ∥’ stands for ‘parallel to’ )
Example 2
AB is a line segment and line l is its perpendicular bisector. If a point P lies on l, show that P is equidistant from A and B.
Solution :
Line l ⊥ AB and passes through C which is the mid-point of AB (see Figure 7.9). We have to show that PA = PB.
Consider ΔPCA and ΔPCB.
Figure 7.9
We have AC = BC (C is the mid-point of AB)
∠PCA = ∠PCB = 90° (Given)
Side PC = Side PC (Common)
So, ΔPCA congruent to ΔPCB (SAS rule)
and so, Side PA = Side PB, as they are corresponding sides of congruent triangles.
Now, let us construct two triangles, whose sides are 4 cm and 5 cm and one of the angles is 50° and this angle is not included in between the equal sides (see Figure 7.10).
Figure 7.10
Are the two triangles congruent?
Notice that the two triangles are not congruent.
Repeat this activity with more pairs of triangles. You will observe that for two triangles to be congruent, it is very important that the equal angles are included between the pairs of equal sides.
So, SAS congruence rule holds but not ASS or SSA rule.
Next, try to construct two triangles in which two angles are 60° and 45° and the side included between these two angles is 4 cm (see Figure 7.11).
Figure 7.11
Cut out these triangles and place one triangle on the other. What do you observe? See that one triangle covers the other completely; that is, the two triangles are congruent. Repeat this activity with some more pairs of triangles. You will observe that equality of two angles and the included side is just sufficient for congruence of triangles.
This result is the Angle-Side-Angle criterion for congruence and is written as ASA criterion. We have verified this criterion in our earlier classes but let us state and prove this result.
Since this result can be proved, it is called a theorem and to prove it, we use the SAS axiom for congruence.
Theorem 7.1 (ASA congruence rule)
Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.
Proof :
We are given two triangles ABC and DEF in which:
∠B = ∠E; ∠C = ∠F; and Side BC = Side eF
We need to prove that ΔABC ≅ ΔDEF
For proving the congruence of the two triangles note that three cases arise.
Case (i) :
Let AB = dE (see Figure 7.12).
Figure 7.12
Now what do you observe? You may observe that
AB = dE (assumed)
∠B = ∠E (Given)
BC = eF (Given)
So, ΔABC ≅ ΔDEF (By SAS rule)
Case (ii) :
Let us assume that AB > dE.
So, we can take a point P on AB such that
PB = dE.
Now consider ΔPBC and ΔDEF (see Figure 7.13).
Fig 7.13
Observe that in ΔPBC and ΔDEF,
PB = dE (By construction)
∠B = ∠E (Given)
BC = eF (Given)
So, we can conclude that:
ΔPBC ≅ ΔDEF, by the SAS axiom for congruence.
Since the triangles are congruent, their corresponding parts will be equal.
So, ∠PCB = ∠DFE
But we are given that
∠ACB = ∠DFE
So, ∠ACB = ∠PCB
Is this possible?
This is possible only if P coincides with A.
or BA = eD
So, ΔABC ≅ ΔDEF (by SAS axiom)
Case (iii) :
If AB < dE,
we can choose a point M on dE such that ME = AB and repeating the arguments as given in Case (ii), we can conclude that Side AB = Side dE and so, ΔABC ≅ΔDEF.
End of proof.
Suppose now in two triangles two pairs of angles and one pair of corresponding sides are equal but the side is not included between the corresponding equal pairs of angles. Are the triangles still congruent? We observe that they are congruent. Can we reason out why?
We know that the sum of the three angles of a triangle is 180°. So, if two pairs of angles are equal, the third pair is also equal (180° − sum of equal angles).
So, two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal. We may call it as the \(AAS\) Congruence Rule.
Now let us perform the following activity :
Draw triangles with angles 40°, 50° and 90°. How many such triangles can you draw?
In fact, we can draw as many triangles as we want with different lengths of sides (see Figure 7.14).
Figure 7.14
Observe that the triangles may or may not be congruent to each other.
So, equality of three angles is not sufficient for congruence of triangles. Therefore, for congruence of triangles out of three equal parts, one has to be a side.
Let us now take some more examples.
Example 3
Line segment AB is parallel to another line segment CD. O is the mid-point of AD (see Figure 7.15).
Fig 7.15
Show that
(i)ΔAOB ≅ ΔDOC
(ii) O is also the mid-point of BC.
Solution :
(i) Consider ΔAOB and ΔDOC.
∠ABO = ∠DCO
(Alternate angles as AB ∥ CD and BC is the transversal) [ AB & CD are parallel]
∠AOB = ∠DOC .. .. (Vertically opposite angles)
OA = OD (Given)
Therefore, ΔAOB ≅ ΔDOC ( AAS rule)
(ii) OB = OC (CPCT)
So, O is the mid-point of BC.
EXERCISE 7.1
Q1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see Figure 7.16).
Figure 7.16
Show that ΔABC ≅ ΔABD. What can you say about BC and BD?
A1. They are equal.
Q2. ABCD is a quadrilateral in which AD = BC and \(∠DAB = ∠CBA\) (see Figure 7.17).
Figure 7.17
Show that CD bisects AB.
Q4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Figure 7.19).
Figure 7.19
Show that ΔABC ≅ ΔCDA.
Q5. Line l is the bisector of an ∠A, and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see Figure 7.20).
Figure 7.20
Show that:
(i)ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Q6. In Figure 7.21,
AC = AE, AB = AD and ∠BAD = ∠EAC.
Figure 7.21
Show that BC = dE.
A6. ∠BAC = ∠DAE
Q7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see Figure 7.22).
Figure 7.22
Show that
(i)ΔDAP ≅ ΔEBP
(ii) AD = bE
Q8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Figure 7.23).
Figure 7.23
Show that:
(i)ΔAMC ≅ ΔBMD
(ii) ∠DBC is a right angle.
(iii)ΔDBC ≅ ΔACB
(iv) CM = ½ AB
7.4 Some Properties of a triangle
NCERT Class 9 Mathematics Textbook for Blind Students made Screen Readable by Professor T K Bansal.
In the above section we studied two criteria for congruence of triangles. Let us now apply these results to study some properties related to a triangle whose both sides are equal.
Perform the activity given below:
Construct a triangle in which two sides are equal, say each = 3.5 cm and the third side = 5 cm (see Figure 7.24).
Figure 7.24
We have done such constructions in our earlier classes.
Do you remember what is such a triangle called?
A triangle in which two sides are equal is called an isosceles triangle. So, ΔABC of Figure 7.24 is an isosceles triangle with AB = AC.
Now, measure ∠B and ∠C. What do you observe?
Repeat this activity with other isosceles triangles with different sides.
We observe that in each such triangle, angles opposite to the equal sides are equal.
This is a very important result and is indeed true for any isosceles triangle.
It can be proved as shown below.
Theorem 7.2 Angles opposite to equal sides of an isosceles triangle are equal.
This result can be proved in many ways. One of the proofs is given here.
Proof :
We are given an isosceles triangle ABC in which AB = AC. We need to prove that ∠B = ∠C.
Let us draw the bisector of ∠A and let D be the point of intersection of this bisector of angle A and BC (see Figure 7.25).
Figure 7.25
In ΔBAD and ΔCAD,
AB = AC (Given)
∠BAD = ∠CAD (By construction)
AD = AD (Common)
So, ΔBAD ≅ ΔCAD (By SAS rule)
So, ∠ABD = ∠ACD, since they are corresponding angles of congruent triangles.
So, ∠B = ∠C
Is the converse also true? That is:
If two angles of any triangle are equal, can we conclude that the sides opposite to them are also equal?
Perform the following activity.
Construct a triangle ABC with BC of any length and ∠B = ∠C = 50°. Draw the bisector of ∠A, and let it intersect BC at D (see Figure 7.26).
Figure 7.26
Cut out the triangle from the sheet of paper and fold it along AD so that vertex C falls on vertex B. What can you say about the sides AC and AB? Observe that AC covers AB completely
So, AC = AB
Repeat this activity with some more triangles. Each time you will observe that the sides opposite to equal angles are equal. So, we have the following:
Theorem 7.3
The sides opposite to equal angles of a triangle are equal.
This is the converse of Theorem 7.2.
We can prove this theorem by ASA congruence rule.
Let us take some examples to apply these results.
Example 4
In triangle ABC, the bisector AD of ∠A is perpendicular to side BC (see Figure 7.27).
Figure 7.27
Show that AB = AC and triangle ABC is isosceles.
Solution :
In ΔABD and ΔACD,
∠BAD = ∠CAD (Given)
AD = AD (Common)
∠ADB = ∠ADC = 90° (Given)
So, ΔABD ≅ ΔACD (ASA rule)
So, AB = AC (CPCT)
or ΔABC is an isosceles triangle.
Example 5
E and F are respectively the mid-points of equal sides AB and AC of a triangle ABC (see Figure 7.28).
Fig 7.28
Show that BF = cE.
Solution :
In ΔABF and in ΔACE,
AB = AC (Given)
∠A = ∠A (Common)
AF = AE (Halves of equal sides)
So, ΔABF ≅ ΔACE (SAS rule)
Therefore, BF = cE (CPCT)
Example 6
In an isosceles triangle ABC with AB = AC, D and E are points on BC such that BE = CD (see Figure 7.29).
Fig 7.29
Show that AD = AE.
Solution :
In ΔABD and ΔACE,
AB = AC (Given) Equation (1)
∠B = ∠C (Angles opposite to equal sides) Equation (2)
Also, bE = CD
So, bE − dE = CD − dE
That is, BD = cE (Equation 3)
So, ΔABD ≅ ΔACE
(Using Equations (1), (2), (3) and SAS rule).
This gives AD = AE (CPCT)
EXERCISE 7.2
Q1. In an isosceles triangle ABC, with Side AB = Side AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that :
(i) OB = OC
(ii) AO bisects ∠A
Q2. In triangle ABC, AD is the perpendicular bisector of BC (see Figure 7.30).
Fig 7.30
Show that ΔABC is an isosceles triangle in which AB = AC.
Q3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Figure 7.31).
Fig 7.31
Show that these altitudes are equal.
Q4. ABC is a triangle in which altitudes bE and CF to sides AC and AB are equal (see Figure 7.32).
Figure 7.32
Show that
(i)ΔABE ≅ ΔACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
Q5. ABC and DBC are two isosceles triangles on the same base BC (see Figure 7.33).
Figure 7.33
Show that ∠ABD = ∠ACD.
Q6.ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Figure 7.34).
Figure 7.34
Show that ∠BCD is a right angle.
A6. ∠BCD = ∠BCA + ∠DCA = ∠B + ∠D
Q7. ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Q8. Show that the angles of an equilateral triangle are 60° each.
7.5 Some More Criteria for Congruence of triangles
NCERT Class 9 Mathematics Textbook for Blind Students made Screen Readable by Professor T K Bansal.
We have seen earlier in this chapter that equality of three angles of one triangle to three angles of the other is not sufficient for the congruence of the two triangles. We may wonder whether equality of three sides of one triangle to three sides of another triangle is enough for congruence of the two triangles. We have already verified in earlier classes that this is indeed true.
To be sure, construct two triangles with sides 4 cm, 3.5 cm and 4.5 cm (see Figure 7.35).
Figure 7.35
Cut them out and place them on each other. What do you observe? They cover each other completely, if the equal sides are placed on each other. So, the triangles are congruent.
Repeat this activity with some more triangles. We arrive at another rule for congruence.
Theorem 7.4 (SSS congruence rule)
If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.
This theorem can be proved using a suitable construction.
We have already seen that in the SAS congruence rule, the pair of equal angles has to be the included ∠between the pairs of corresponding pair of equal sides and if this is not so, the two triangles may not be congruent.
Perform this activity:
Construct two right angled triangles with hypotenuse = 5 cm and one side = 4 cm each (see Figure 7.36).
Figure 7.36
Cut them out and place one triangle over the other with equal side placed on each other. Turn the triangles, if necessary. What do you observe?
The two triangles cover each other completely and so they are congruent. Repeat this activity with some other pairs of right triangles. What do we observe?
We will find that two right triangles are congruent if one pair of sides and the hypotenuse are equal. We have verified this in our earlier classes.
Note that, the right angle is not the included angle in this case.
So, we arrive at the following congruence rule:
Theorem 7.5 (RHS congruence rule)
If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.
Note that RHS stands for Right ∠- Hypotenuse - Side.
Let us now take some examples.
Example 7
AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (see Figure 7.37).
Figure 7.37
Show that the line PQ is the perpendicular bisector of AB.
Solution :
We are given that PA = PB and QA = QB.
We are to show that PQ ⊥ AB and PQ bisects AB.
Let PQ intersect AB at C.
Can you think of two congruent triangles in this figure?
Let us take ΔPAQ and ΔPBQ.
In these triangles,
AP = BP (Given)
AQ = BQ (Given)
PQ = PQ (Common)
So, ΔPAQ ≅ ΔPBQ (SSS rule)
Therefore, ∠APQ = ∠BPQ (CPCT).
Now let us consider ΔPAC and ΔPBC.
We have :
AP = BP (Given)
angle APC = ∠BPC (angle APQ = ∠BPQ proved above)
PC = PC (Common)
So, ΔPAC ≅ ΔPBC (SAS rule)
Therefore, AC = BC (CPCT) (1)
and ∠ACP = ∠BCP (CPCT)
Also, ∠ACP + ∠BCP = 180° (Linear pair)
So, 2 × ∠ACP = 180°
or ∠ACP = 90° (2)
From (1) and (2), we can easily conclude that PQ is the perpendicular bisector of AB.
[Note that, without showing the congruence of ΔPAQ and ΔPBQ, we cannot show that ΔPAC ≅ ΔPBC even though
AP = BP (Given)
PC = PC (Common)
and ∠PAC = ∠PBC (Angles opposite to equal sides in ΔAPB)
It is because these results give us SSA rule which is not always valid or true for congruence of triangles. Also, the angle is not included between the equal pairs of sides.]
Let us take some more examples.
Example 8
P is a point equidistant from two lines l and m intersecting at point A (see Figure 7.38).
Figure 7.38
Show that the line AP bisects the ∠between them.
Solution :
We are given that lines l and m intersect with each other at A. Let PB ⊥ l,
PC ⊥ m. It is given that PB = PC.
We are to show that ∠PAB = ∠PAC.
Let us consider ΔPAB and ΔPAC.
In these two triangles,
PB = PC (Given)
∠PBA = ∠PCA = 90° (Given)
PA = PA (Common)
So, ΔPAB ≅ ΔPAC (RHS rule)
So, ∠PAB = ∠PAC (CPCT)
Note that this result is the converse of the result proved in Q5 of Exercise 7.1.
EXERCISE 7.3
Q1. ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Figure 7.39).
Figure 7.39
If AD is extended to intersect BC at P, show that
(i)ΔABD ≅ ΔACD
(ii)ΔABP ≅ ΔACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.
Q2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A.
Q3. Both sides AB and BC; and median aM of one triangle ABC are respectively equal to sides PQ and QR, and median PN of ΔPQR (see Figure 7.40).
Fig 7.40
Show that:
(i)ΔABM ≅ ΔPQN
(ii)ΔABC ≅ ΔPQR
A3.
(ii) From (i), ∠ABM = ∠PQN
Q4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the ΔABC is isosceles.
Q5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
7.6 Inequalities in a triangle
NCERT Class 9 Mathematics Textbook for Blind Students made Screen Readable by Professor T K Bansal.
So far, we have been mainly studying the equality of sides and angles of a triangle or of triangles. Sometimes, we do come across unequal objects, we need to compare them as well.
For example, line segment AB is greater in length as compared to line segment CD in Figure 7.41 (i) and ∠A is greater than ∠B in Fig 7.41 (ii).
Fig 7.41 (i)