This topic gives an overview of the theorems;
Let us apply the results of congruence of triangles to study some properties related to a triangle whose two sidesare equal.
Perform the activity given below:
Construct a triangle in which two sides are equal, say each equal to 3.5 cm and the third side equal to 5 cm. You have done such constructions in earlier classes.
A triangle in which two sides are equal is called an isosceles triangle. So, Δ ABC is an isosceles triangle with AB = AC. Now, measure ∠ B and ∠ C. Repeat this activity with other isosceles triangles with different sides. You may observe that in each such triangle, the 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 : 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.
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 ∠ A and BC.
In Δ BAD and Δ CAD,
AB = AC (Given)
∠BAD = ∠CAD (By construction)
AD = AD (Common)
So, Δ BAD ≅ Δ CAD (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 .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 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 : The sides opposite to equal angles of a triangle are equal
This is the converse of Theorem 7.2. You can prove this theorem by ASA congruence rule.Let us take some examples to apply these results.
Example: In Δ ABC, the bisector AD of ∠ A is perpendicular to side BC.Show that AB = AC and Δ ABC is isosceles.
Solution : In ΔABD and ΔACD,
∠BAD = ∠CAD (Given)
AD = AD (Common)
∠ADB = ∠ADC = 90° (Given)
So, Δ ABD ≅ Δ ACD (ASA rule)
So, AB = AC (CPCT) or, Δ ABC is an isosceles triangle.
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. 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.
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.
Cut them out and place one triangle over the other with equal side placed on each other. Turn the triangles, if necessary. The two triangles cover each other completely and so they are congruent.
Repeat this activity with other pairs of right triangles. 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:
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.
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. Show that the line PQ is the perpendicular bisector of AB.
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.
Let us take Δ PAQ and Δ PBQ. In these triangles,
AP = BP (Given)
AQ = BQ (Given)
PQ = PQ (Common)
So, Δ PAQ ≅ Δ PBQ (SSS rule)
Therefore, ∠APQ = ∠BPQ (CPCT).
Now let us consider Δ PAC and Δ PBC.
You have : AP = BP (Given)
∠APC = ∠BPC (∠APQ = ∠BPQ proved above)
PC = PC (Common)
So, Δ PAC ≅ Δ PBC (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), you can easily conclude that PQ is the perpendicular bisector of AB. [Note that, without showing the congruence of Δ PAQ and Δ PBQ, you cannot show that Δ PAC ≅ Δ PBC 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.]
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