`color{red} ♦` Some Properties of a Triangle

In the above section you have studied two criteria for congruence of triangles. Let us now apply these results to study some properties related to a triangle whose two sides are 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` (see Fig. `7.24`). You have done such constructions in 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, `Delta ABC` of Fig. `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.

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.

` color{blue} text(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.

` color{blue} text(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 `∠ A` and `BC` (see Fig. 7.25).

In `Delta BAD` and `Delta CAD`,

`AB = AC` (Given)

`∠ BAD = ∠ CAD` (By construction)

`AD = AD` (Common)

So, `Delta BAD ≅ Delta 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` (see Fig. 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 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:

` color{blue} text ( Theorem 7.3)` :

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.

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` (see Fig. `7.24`). You have done such constructions in 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, `Delta ABC` of Fig. `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.

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.

` color{blue} text(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.

` color{blue} text(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 `∠ A` and `BC` (see Fig. 7.25).

In `Delta BAD` and `Delta CAD`,

`AB = AC` (Given)

`∠ BAD = ∠ CAD` (By construction)

`AD = AD` (Common)

So, `Delta BAD ≅ Delta 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` (see Fig. 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 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:

` color{blue} text ( Theorem 7.3)` :

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.

Q 3210178010

In `Delta ABC`, the bisector `AD` of `∠ A` is perpendicular to side `BC` (see Fig. `7.27`). Show that `AB = AC` and `Delta ABC` is isosceles.

Class 9 Chapter 7 Example 4

Class 9 Chapter 7 Example 4

In `Delta ABD` and `Delta ACD`,

`∠ BAD = ∠ CAD` (Given)

`AD = AD` (Common)

`∠ ADB = ∠ ADC = 90°` (Given)

So, `Delta ABD ≅ Delta ACD` (ASA rule)

So, `AB = AC (CPCT)`

or, `Delta ABC` is an isosceles triangle.

Q 3220178011

`E` and `F` are respectively the mid-points of equal sides `AB` and `AC` of `Delta ABC` (see Fig. 7.28). Show that `BF = CE`.

Class 9 Chapter 7 Example 5

Class 9 Chapter 7 Example 5

In `Delta ABF` and `Delta ACE`,

`AB = AC` (Given)

`∠ A = ∠ A` (Common)

`AF = AE` (Halves of equal sides)

So, `Delta ABF ≅ Delta ACE` (SAS rule)

Therefore, `BF = CE` (CPCT)

Q 3230178012

In an isosceles triangle `ABC` with `AB = AC, D` and `E` are points on `BC` such that `BE = CD` (see Fig. 7.29). Show that `AD = AE`.

Class 9 Chapter 7 Example 6

Class 9 Chapter 7 Example 6

In `Delta ABD` and `Delta ACE`,

`AB = AC` (Given) (1)

`∠ B = ∠ C`

(Angles opposite to equal sides) (2)

Also, `BE = CD`

So, `BE – DE = CD – DE`

That is, `BD = CE` (3)

So, `Delta ABD ≅ Delta ACE`

(Using (1), (2), (3) and SAS rule).

This gives `AD = A E`(CPCT)