Criteria for Congruence of Triangles

We make use of triangular structures and patterns frequently in day-to-day life. So, it is rewarding to find out when two triangular shapes will be congruent. If you have two triangles drawn in your notebook and want to verify if they are congruent, you cannot everytime cut out one of them and use method of superposition. Instead, if we can judge congruency in terms of approrpriate measures, it would be quite useful. Let us try to do this.

Appu and Tippu play a game. Appu has drawn a triangle ABC (Fig 7.8) and has noted the length of each of its sides and measure of each of its angles. Tippu has not seen it. Appu challenges Tippu if he can draw a copy of his ΔABC based on bits of information that Appu would give. Tippu attempts to draw a triangle congruent to ΔABC, using the information provided by Appu. The game starts. Carefully observe their conversation and their games.

Appu : One side of ΔABC is 5.5 cm.

Tippu : With this information, I can draw any number of triangles (Fig 7.9) but they need not be copies of ΔABC. The triangle I draw may be obtuse-angled or right-angled or acute-angled. For example, here are a few.

I have used some arbitrary lengths for other sides. This gives me many triangles with length of base 5.5 cm.

So, giving only one side-length will not help me to produce a copy of ΔABC.

Appu : Okay. I will give you the length of one more side. Take two sides of ΔABC to be of lengths 5.5 cm and 3.4 cm.

Tippu : Even this will not be sufficient for the purpose. I can draw several triangles (Fig 7.10) with the given information which may not be copies of ΔABC. Here are a few to support my argument:

One cannot draw an exact copy of your triangle, if only the lengths of two sides are given.

Appu : Alright. Let me give the lengths of all the three sides. In ΔABC, I have AB = 5cm, BC = 5.5 cm and AC = 3.4 cm.

Tippu : I think it should be possible. Let me try now. First I draw a rough figure so that I can remember the lengths easily.

I draw BC with length 5.5 cm.

With B as centre, I draw an arc of radius 5 cm. The point A has to be somewhere on this arc. With C as centre, I draw an arc of radius 3.4 cm. The point A has to be on this arc also.

So, A lies on both the arcs drawn. This means A is the point of intersection of the arcs.

I know now the positions of points A, B and C. Aha! I can join them and get ΔABC

(Fig 7.11).

Appu : Excellent. So, to draw a copy of a given ΔABC (i.e., to draw a triangle congruent to ΔABC), we need the lengths of three sides. Shall we call this condition as side-side-side criterion?

Tippu : Why not we call it SSS criterion, to be short?

EXAMPLE 2

In triangles ABC and PQR, AB = 3.5 cm, BC = 7.1 cm, AC = 5 cm, PQ = 7.1 cm, QR = 5 cm and PR = 3.5 cm. Examine whether the two triangles are congruent or not. If yes, write the congruence relation in symbolic form.

Here, AB = PR (= 3.5 cm),

and BC = PQ ( = 7.1 cm) AC = QR (= 5 cm)

This shows that the three sides of one triangle are equal to the three sides of the other triangle. So, by SSS congruence rule, the two triangles are congruent. From the above three equality relations, it can be easily seen

that A ↔ R, B ↔ P and C ↔ Q.

So, we have ΔABC ≅ ΔRPQ

Important note:

The order of the letters in the names of congruent triangles displays the corresponding relationships. Thus, when you write ΔABC ≅ ΔRPQ, you would know that A lies on R, B on P, C on Q, AB along RP , BC along PQ and AC along RQ .

EXAMPLE 3 In Fig 7.13, AD = CD and AB = CB.

(i) State the three pairs of equal parts in ΔABD and ΔCBD. (ii) Is ΔABD ≅ ΔCBD? Why or why not?

(iii) Does BD bisect ∠ABC? Give reasons.

SOLUTION

(i) In ΔABD and ΔCBD, the three pairs of equal parts are as given below:

AB = CB (Given) AD = CD (Given) and BD = BD (Common in both)

(ii) From (i) above, ΔABD ≅ ΔCBD (By SSS congruence rule)

(iii) ∠ABD = ∠CBD (Corresponding parts of congruent triangles)

So, BD bisects ∠ABC.

ABC is an isosceles triangle with AB = AC (Fig 7.17). A

Take a trace-copy of ΔABC and also name it as ΔABC.

(i) State the three pairs of equal parts in ΔABC and ΔACB.

(ii) Is ΔABC ≅ ΔACB? Why or why not?

(iii) Is ∠B = ∠C ? Why or why not?

Appu and Tippu now turn to playing the game with a slight modification.

Appu : Let me now change the rules of the triangle-copying game.

Tippu : Right, go ahead. B CFig 7.17

Appu : You have already found that giving the length of only one side is useless.

Tippu : Of course, yes.

Appu : In that case, let me tell that in ΔABC, one side is 5.5 cm and one angle is 65°.

Tippu : This again is not sufficient for the job. I can find many triangles satisfying your information, but are not copies of ΔABC. For example, I have given here some of them

(Fig 7.18):

Appu : So, what shall we do?

Tippu : More information is needed.

Appu : Then, let me modify my earlier statement. In ΔABC, the length of two sides are

5.5 cm and 3.4 cm, and the angle between these two sides is 65°.

Tippu : This information should help me. Let me try. I draw first BC of length 5.5. cm

[Fig 7.19 (i)]. Now I make 65° at C [Fig 7.19 (ii)].

Yes, I got it, A must be 3.4 cm away from C along this angular line through C.

I draw an arc of 3.4 cm with C as centre. It cuts the 65° line at A. Now, I join AB and get ΔABC [Fig 7.19(iii)].

Appu : You have used side-angle-side, where the angle is ‘included’between the sides!

Tippu : Yes. How shall we name this criterion? Appu : It is SAS criterion. Do you follow it? Tippu : Yes, of course.

EXAMPLE 4

Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not, by using SAS congruence rule. If the triangles are congruent, write them in symbolic form.

ΔABC ΔDEF

(a) AB = 7 cm, BC = 5 cm, ∠B = 50° DE = 5 cm, EF = 7 cm, ∠E = 50°

(b) AB = 4.5 cm, AC = 4 cm, ∠A = 60° DE = 4 cm, FD = 4.5 cm, ∠D = 55°

(c) BC = 6 cm, AC = 4 cm, ∠B = 35° DF = 4 cm, EF = 6 cm, ∠E = 35°

(It will be always helpful to draw a rough figure, mark the measurements and then probe the question).

(a) Here, AB = EF ( = 7 cm), BC = DE ( = 5 cm) and

included ∠B = included ∠E ( = 50°). Also, A ↔ F B ↔ E and C ↔ D. Therefore, ΔABC ≅ ΔFED (By SAS congruence rule) (Fig 7.20)

(b) Here, AB = FD and AC = DE (Fig 7.21).

But included ∠A ≠ included ∠D. So, we cannot say that the triangles are congruent.

(c) Here, BC = EF, AC = DF and ∠B = ∠E.

But ∠B is not the included angle between the sides AC and BC.

Similarly, ∠E is not the included angle between the sides EF and DF. D

So, SAS congruence rule cannot be applied and we cannot conclude

that the two triangles are congruent.

EXAMPLE 5

In Fig 7.23, AB = AC and AD is the bisector of ∠BAC. (i) State three pairs of equal parts in triangles ADB and ADC.

(ii) Is ΔADB ≅ ΔADC? Give reasons.

(iii) Is ∠B = ∠C? Give reasons.

In Fig 7.23, AB = AC and AD is the bisector of ∠BAC.

(i) State three pairs of equal parts in triangles ADB and ADC.

(ii) Is ΔADB ≅ ΔADC? Give reasons.

(iii) Is ∠B = ∠C? Give reasons.

(i) The three pairs of equal parts are as follows:

AB = AC (Given)

∠BAD = ∠CAD (AD bisects ∠BAC) and AD = AD (common)

(ii) Yes, ΔADB ≅ ΔADC (By SAS congruence rule)

(iii) ∠B = ∠C (Corresponding parts of congruent triangles)

Can you draw Appu’s triangle, if you know

(i) only one of its angles? (ii) only two of its angles? (iii) two angles and any one side?

(iv) two angles and the side included between them?

Attempts to solve the above questions lead us to the following criterion:

By applyingASAcongruence rule, it is to be established that ΔABC ≅ ΔQRP

and it is given that BC = RP. What additional information is needed to

establish the congruence?

For ASA congruence rule, we need the two angles between which the two sides BC and RP are included. So, the additional information is

as follows:

∠B = ∠R

and ∠C = ∠P

EXAMPLE 7 In Fig 7.26, can you use ASA congruence rule and conclude that ΔAOC ≅ ΔBOD?

SOLUTION In the two triangles AOC and BOD, ∠C = ∠D (each 70° )

Also, ∠AOC = ∠BOD = 30° (vertically opposite angles)

So, ∠A of ΔAOC = 180° – (70° + 30°) = 80°

SOLUTION In the two triangles AOC and BOD, ∠C = ∠D (each 70° )

Also, ∠AOC = ∠BOD = 30° (vertically opposite angles)

So, ∠A of ΔAOC = 180° – (70° + 30°) = 80°

(using angle sum property of a triangle) Similarly, ∠B of ΔBOD = 180° – (70° + 30°) = 80°

Thus, we have ∠A = ∠B, AC = BD and ∠C = ∠D

Now, side AC is between ∠A and ∠C and side BD is between ∠B and ∠D.

So, by ASA congruence rule, ΔAOC ≅ ΔBOD.

Given two angles of a triangle, you can always find the third angle of the triangle. So, whenever, two angles and one side of one triangle are equal to the corresponding two angles and one side of another triangle, you may convert it into ‘two angles and the included side’ form of congruence and then apply the ASA congruence rule.

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