Class 9 Some More Criteria for Congruence of Triangles

### Topic covered

color{red} ♦Some More Criteria for Congruence of Triangles

### Some More Criteria for Congruence of Triangles

You 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.

You may wonder whether equality of three sides of one triangle to three sides of another triangle is enough for congruence of the two triangles. You 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 Fig. 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.

color{blue} text( 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.

You have already seen that in the SAS congruence rule, the pair of equal angles has to be the included angle 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 equal to 5 cm and one side equal to 4 cm each (see Fig. 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 other pairs of right triangles. What do you observe?

You will find that two right triangles are congruent if one pair of sides and the hypotenuse are equal. You have verified this in earlier classes.

Note that, the right angle is not the included angle in this case.

So, you arrive at the following congruence rule:

 color{ blue } text( 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 angle - Hypotenuse - Side.

Let us now take some examples.
Q 3250178014

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 Fig. 7.37). Show that the line PQ is the perpendicular bisector of AB.
Class 9 Chapter 7 Example 7
Solution:

You are given that PA = PB and

QA = QB and you 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 Delta PAQ and Delta PBQ.

In these triangles,

AP = BP (Given)

AQ = BQ (Given)

PQ = PQ (Common)

So, Delta PAQ ≅ Delta PBQ (SSS rule)

Therefore, ∠ APQ = ∠ BPQ (CPCT).

Now let us consider Delta PAC and Delta PBC.

You have : AP = BP (Given)

∠ APC = ∠ BPC (∠ APQ = ∠ BPQ proved above)

PC = PC (Common)

So, Delta PAC ≅ Delta PBC (SAS rule)

Therefore, AC = BC (CPCT) (1)

and ∠ ACP = ∠ BCP (CPCT)

Also, ∠ ACP + ∠ BCP = 180°

So, 2 ∠ ACP = 180°

or, ∠ ACP = 90° (2)

From (1) and (2), you can easily conclude that PQ is the perpendicular bisector of AB.
[Note that, without showing the congruence of Delta PAQ and Delta PBQ, you cannot show
that Delta PAC ≅ Delta PBC even though AP = BP (Given)

PC = PC (Common)

and ∠ PAC = ∠ PBC (Angles opposite to equal sides in
Delta APB)

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.]
Q 3200178018

P is a point equidistant from two lines l and m intersecting at point A (see Fig. 7.38). Show that the line AP bisects the angle between them.
Class 9 Chapter 7 Example 8
Solution:

You are given that lines l and m intersect each other at A. Let PB ⊥ l,

PC ⊥ m. It is given that PB = PC.

You are to show that ∠ PAB = ∠ PAC.

Let us consider D PAB and D PAC. In these two
triangles,

PB = PC (Given)

∠ PBA = ∠ PCA = 90° (Given)

PA = PA (Common)

So, Delta PAB ≅ Delta PAC (RHS rule)

So, ∠ PAB = ∠ PAC (CPCT)