NCERT Class 9 Maths Solutions: Triangles

Question:

Properties of isosceles triangles, angle bisectors, congruent triangles (SSS, SAS, ASA criteria), and the fact that angles opposite equal sides are equal.


Part 1: OB = OC
Given that triangle ABC is isosceles with AB = AC.
This implies that ∠B = ∠C (angles opposite equal sides are equal).
Given that BO bisects ∠B and CO bisects ∠C.
Therefore, ∠OBC = ∠B / 2 and ∠OCB = ∠C / 2.
Since ∠B = ∠C, it follows that ∠B / 2 = ∠C / 2, which means ∠OBC = ∠OCB.
Now consider triangle OBC. Since ∠OBC = ∠OCB, triangle OBC is an isosceles triangle with OB = OC (sides opposite equal angles are equal).

Part 2: AO bisects ∠A
Given that triangle ABC is isosceles with AB = AC.
Given that BO and CO are angle bisectors of ∠B and ∠C, intersecting at O.
We have already shown in Part 1 that OB = OC.
Now consider triangles ABO and ACO.
AB = AC (given)
∠OAB = ∠OAC (This is what we need to prove. We can’t assume it.)
∠OBA = ∠OCA (from Part 1, ∠OBC = ∠OCB, and since BO and CO bisect ∠B and ∠C, this implies ∠ABO = ∠ACO if we consider the whole angle. However, it’s more direct to use the ASA congruence criteria.)

Let’s re-approach Part 2 using congruence:
Consider triangles ABO and ACO.
AB = AC (given)
∠ABO = ∠ACO (Since ∠B = ∠C and BO and CO are angle bisectors, ∠ABO = ∠B/2 and ∠ACO = ∠C/2, hence ∠ABO = ∠ACO)
BO = CO (proven in Part 1)

Using the SAS (Side-Angle-Side) congruence criterion for triangles ABO and ACO:
AB = AC (Side)
∠ABO = ∠ACO (Angle)
BO = CO (Side)
Therefore, triangle ABO is congruent to triangle ACO (SAS congruence).

By the CPCTC (Corresponding Parts of Congruent Triangles are Congruent):
∠BAO = ∠CAO
This means that AO bisects ∠A.

Question:

Congruence of Triangles: Two triangles are congruent if their corresponding sides and angles are equal. The common congruence criteria are SSS, SAS, ASA, AAS, and RHS.
Properties of Isosceles Triangles: In an isosceles triangle, the angles opposite the equal sides are equal.


We are given an isosceles triangle ABC where AB = AC. We are also given that AP is perpendicular to BC, which means ∠APB = ∠APC = 90°. We need to show that ∠B = ∠C.

To show that ∠B = ∠C, we can prove that triangle ABP is congruent to triangle ACP.

Consider triangles ABP and ACP:
1. AB = AC (Given, as ABC is an isosceles triangle)
2. AP = AP (Common side)
3. ∠APB = ∠APC (Given, as AP ⊥ BC)

Since we have two sides and the included angle equal, we can use the SAS (Side-Angle-Side) congruence criterion. However, the angle given is ∠APB and ∠APC, which are right angles. This means we have two sides and a right angle. We can also consider the RHS (Right-angle-Hypotenuse-Side) congruence criterion if we establish that the hypotenuses are equal and one pair of corresponding sides are equal.

Let’s re-examine the given information and what we can use. We have:
1. AB = AC (Given)
2. AP = AP (Common side)
3. ∠APB = ∠APC = 90° (Given, AP ⊥ BC)

In triangle ABP, AB is the hypotenuse. In triangle ACP, AC is the hypotenuse. Since AB = AC, the hypotenuses are equal. AP is a common side and it is also a side for both triangles. Therefore, by the RHS congruence criterion:

Triangle ABP ≅ Triangle ACP (RHS congruence)

Now, since the triangles are congruent, their corresponding angles are equal.
Therefore, ∠B = ∠C.

Question:

RHS Congruence Rule: If in two right-angled triangles, the hypotenuse and one side of one triangle are equal to the hypotenuse and the corresponding side of the other triangle, then the two triangles are congruent.
Isosceles Triangle: A triangle with at least two sides of equal length.


Given: BE and CF are two equal altitudes of a triangle ABC. This means BE = CF.
Also, since BE and CF are altitudes, they are perpendicular to the sides AC and AB respectively.
Therefore, angle BEC = 90 degrees and angle CFB = 90 degrees.

We need to prove that triangle ABC is isosceles, which means we need to show that AB = AC.

Consider triangles BEC and CFB.
1. Hypotenuse: BC is common to both triangles BEC and CFB. So, BC = BC.
2. Right Angle: Angle BEC = Angle CFB = 90 degrees (given that BE and CF are altitudes).
3. Side: BE = CF (given that they are equal altitudes).

By the RHS congruence rule, triangle BEC is congruent to triangle CFB.
(Hypotenuse-Leg Congruence Rule, as hypotenuse BC is common and one leg BE = CF)

Since triangle BEC is congruent to triangle CFB, their corresponding parts are equal.
Therefore, angle BCE = angle CBF.
In triangle ABC, angle BCE is the same as angle ACB, and angle CBF is the same as angle ABC.
So, angle ACB = angle ABC.

In a triangle, if two angles are equal, then the sides opposite to these angles are also equal.
The side opposite to angle ABC is AC.
The side opposite to angle ACB is AB.
Therefore, AC = AB.

Since two sides of triangle ABC (AB and AC) are equal, triangle ABC is an isosceles triangle.

Question:

Congruence of triangles. Specifically, RHS (Right angle-Hypotenuse-Side) congruence rule and SAS (Side-Angle-Side) congruence rule. Properties of isosceles triangles: equal sides opposite equal angles.


We are given an isosceles triangle ABC where AB = AC, and AD is an altitude. This means AD is perpendicular to BC, so ∠ADB = ∠ADC = 90°.

Part 1: AD bisects BC
To show that AD bisects BC, we need to prove that BD = DC. We can do this by proving that triangle ABD is congruent to triangle ACD.

Consider triangles ABD and ACD.
1. ∠ADB = ∠ADC (Both are 90° because AD is an altitude).
2. AB = AC (Given, as ABC is an isosceles triangle).
3. AD is common to both triangles.

Using the RHS congruence rule (Right angle, Hypotenuse, Side), we can conclude that triangle ABD is congruent to triangle ACD.
Since the triangles are congruent, their corresponding parts are equal. Therefore, BD = DC. This means AD bisects BC.

Part 2: AD bisects ∠A
To show that AD bisects ∠A, we need to prove that ∠BAD = ∠CAD. We can again use the congruence of triangles ABD and ACD that we proved in Part 1.

Since triangle ABD is congruent to triangle ACD (as proved by RHS congruence rule), their corresponding angles are equal.
Therefore, ∠BAD = ∠CAD. This means AD bisects ∠A.

Question:

Properties of a right-angled triangle: The sum of angles in any triangle is 180°. In a right-angled triangle, one angle is 90°.
Isosceles triangle property: In an isosceles triangle, the angles opposite to the equal sides are equal.


Given that ABC is a right-angled triangle with ∠A = 90°.
This means that the sum of the other two angles, ∠B and ∠C, must be 90° (since the total sum of angles in a triangle is 180°, and 180° – 90° = 90°).
So, ∠B + ∠C = 90°.

We are also given that AB = AC. This indicates that triangle ABC is an isosceles triangle.
In an isosceles triangle, the angles opposite the equal sides are equal.
The angle opposite to side AB is ∠C.
The angle opposite to side AC is ∠B.
Therefore, ∠B = ∠C.

Now we have two equations:
1. ∠B + ∠C = 90°
2. ∠B = ∠C

Substitute equation (2) into equation (1):
∠B + ∠B = 90°
2∠B = 90°
∠B = 90° / 2
∠B = 45°

Since ∠B = ∠C, then ∠C = 45°.

Thus, ∠B = 45° and ∠C = 45°.

Question:

Properties of equilateral triangles: All sides are equal, and all angles are equal.
Sum of angles in a triangle: The sum of the interior angles of any triangle is always 180°.


Let the equilateral triangle be denoted as ABC.
By definition, in an equilateral triangle, all sides are equal in length: AB = BC = CA.
A fundamental property of triangles states that angles opposite equal sides are equal. Since all sides of triangle ABC are equal, all its angles must also be equal. Let each angle be represented by x.
So, angle A = angle B = angle C = x.
The sum of the interior angles of any triangle is 180°.
Therefore, in triangle ABC:
angle A + angle B + angle C = 180°
Substituting x for each angle:
x + x + x = 180°
3x = 180°
To find the value of x, divide both sides of the equation by 3:
x = 180° / 3
x = 60°
Thus, each angle of an equilateral triangle is 60°.