You have studied about triangles and their various properties in your earlier classes. You know that a closed figure formed by three intersecting lines is called a triangle. (‘Tri’ means ‘three’). A triangle has three sides, three angles and three vertices. For
example, in triangle ABC, denoted as Δ ABC (see Fig. 7.1); AB, BC, CA are the three sides, ∠ A, ∠ B, ∠ C are the three angles and A, B, C are three vertices. In Chapter 6, you have also studied some properties of triangles. In this chapter, you will study in detailsabout the congruence of triangles, rules of congruence, some more properties of triangles and inequalities in a triangle. You have already verified most of these properties in earlier classes. We will now prove some of them.
You must have observed that two copies of your photographs of the same size are identical. Similarly, two bangles of the same size, two ATM cards issued by the same bank are identical. You may recall that on placing a one rupee coin on another minted in the same year, they cover each other completely.
Do you 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).
Now, draw two circles of the same radius and place one on the other. What do you observe? They cover each other completely and we call them as congruent circles.
Repeat this activity by placing one square on the other with sides of the same measure (see Fig. 7.2) or by placing two equilateral triangles of equal sides on each other. You will observe that the squares are congruent to each other and so are the equilateral triangles
You may wonder why we are studying congruence. You all must have seen the ice tray in your refrigerator. Observe that the moulds for making ice are all congruent. The cast used for moulding in the tray also has congruent depressions (may be 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, you may find it difficult to replace the refill in your pen by a new one and this is so when the new refill is not of the same size as the one you want to remove. Obviously, if the two refills are identical or congruent, the new refill fits.
So, you 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) :
The large squares in Fig. 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.
You 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 triangle ABC in Fig. 7.4 (i)?
Cut out each of these triangles from Fig. 7.4 (ii) to (v) and turn them around and try to cover Δ ABC. Observe that triangles in Fig. 7.4 (ii), (iii) and (iv) are congruent to Δ ABC while Δ TSU of Fig 7.4 (v) is not congruent to Δ ABC.
If Δ PQR is congruent to Δ ABC, we write Δ PQR ≅ Δ ABC.
Notice that when Δ PQR ≅ Δ ABC, then sides of Δ PQR fall on corresponding equal sides of Δ ABC and so is the case for the angles.
That is, PQ covers AB, QR covers BC and RP covers CA; ∠ 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
Note that under this correspondence, Δ PQR ≅ Δ ABC; but it will not be correct to write ΔQRP ≅ Δ ABC.
Similarly, for Fig. 7.4 (iii),
FD ↔ AB, DE ↔ BC and EF ↔ CA
and F ↔ A, D ↔ B and E ↔ C
So, Δ FDE ≅ Δ ABC but writing Δ DEF ≅ Δ ABC is not correct.
Give the correspondence between the triangle in Fig. 7.4 (iv) and Δ ABC.
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
In earlier classes, you have learnt four 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 Fig. 7.5).
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 are also equal?
In Fig 7.7, BC = QR, ∠ B = ∠ Q and also, AB = PQ. Now, what can you say about congruence of Δ ABC and Δ PQR?
Recall from your earlier classes that, in this case, the two triangles are congruent. Verify this for Δ ABC and Δ PQR in Fig. 7.7.
Repeat this activity with other pairs of triangles. Do you observe that the equality of two sides and the included angle is enough for the congruence of triangles? Yes, it is enough.
This is the first criterion for congruence of triangles.
Axiom (SAS congruence rule) : Two triangles are congruent if two sides and the included angle of one triangle are equal to the two 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 Fig. 7.8, OA = OB and OD = OC. Show that
(i) Δ AOD ≅ Δ BOC and (ii) AD || BC
Solution : (i) You may observe that in Δ AOD and Δ BOC,
Also, since ∠ AOD and ∠ BOC form a pair of vertically opposite angles, we have
∠ AOD = ∠ BOC.
So, Δ AOD ≅ Δ BOC (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, AD || BC.
Example : 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 Fig. 7.9). You have to
show that PA = PB. Consider Δ PCA and Δ PCB.
We have AC = BC (C is the mid-point of AB)
∠ PCA = ∠ PCB = 90° (Given)
PC = PC (Common)
So, Δ PCA ≅ Δ PCB (SAS rule)
and so, PA = 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 Fig. 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 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 the two triangles in which two angles are 60° and 45° and the side included between these angles is 4 cm (see Fig. 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 more pairs of triangles. You will observe that equality of two angles and the included side is sufficient for congruence of triangles.
This result is the Angle-Side-Angle criterion for congruence and is written as ASA criterion. You have verified this criterion in 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 (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 BC = EF
We need to prove that Δ ABC ≅ Δ DEF
For proving the congruence of the two triangles see that three cases arise.
Case (i) : Let AB = DE (see Fig. 7.12).
Now what do you observe? You may observe that
AB = DE (Assumed)
∠ B = ∠ E (Given)
BC = EF (Given)
So, Δ ABC ≅ Δ DEF (By SAS rule)
Case (ii) : Let if possible AB > DE. So, we can take a point P on AB such that PB = DE. Now consider Δ PBC and Δ DEF (see Fig. 7.13).
Observe that in Δ PBC and Δ DEF,
PB = DE (By construction)
∠ B = ∠ E (Given)
BC = EF (Given)
So, we can conclude that:
Δ PBC ≅ Δ DEF, 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, Δ ABC ≅ Δ DEF (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 AB = DE and so,
Δ ABC ≅ Δ DEF.
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? You will observe that they are congruent. Can you reason out why?
You 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, you can draw as many triangles as you want with different lengths of sides (see Fig. 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 : Line-segment AB is parallel to another line-segment CD. O is themid-point of AD (see Fig. 7.15). Show that (i) ΔAOB ≅ ΔDOC (ii) O is also the mid-point of BC.
Solution : (i) Consider Δ AOB and Δ DOC.
∠ ABO = ∠ DCO
(Alternate angles as AB || CD and BC is the transversal)
∠ 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.
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