GSEB Class 9 Maths Solutions Chapter 7 Triangles Exercise 7.2

Question 1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that.
1. OB = OC
2. AO bisects ∠A.

Answer:
**1. To show OB = OC:**
We are given `\( \triangle ABC \)` is an isosceles triangle with `\( AB = AC \)`.
In `\( \triangle ABC \)`, since `\( AB = AC \)`, the angles opposite to these equal sides are also equal.
So, `\( \angle B = \angle C \)`.
Multiplying both sides by `\( \frac{1}{2} \)`, we get `\( \frac{1}{2} \angle B = \frac{1}{2} \angle C \)`.
Since OB and OC are angle bisectors of `\( \angle B \)` and `\( \angle C \)` respectively, we have `\( \angle OBC = \frac{1}{2} \angle B \)` and `\( \angle OCB = \frac{1}{2} \angle C \)`.
Therefore, `\( \angle OBC = \angle OCB \)`.
In `\( \triangle OBC \)`, since `\( \angle OBC = \angle OCB \)`, the sides opposite to these equal angles must also be equal.
Hence, `\( OB = OC \)`.

**2. To show AO bisects ∠A:**
Consider `\( \triangle AOB \)` and `\( \triangle AOC \)`.
We know `\( AB = AC \)` (given).
We have proved `\( OB = OC \)` (from part 1).
Also, `\( AO = AO \)` (common side).
By SSS (Side-Side-Side) congruence rule, `\( \triangle AOB \cong \triangle AOC \)`.
Since the triangles are congruent, their corresponding parts are equal (CPCT).
Therefore, `\( \angle OAB = \angle OAC \)`.
This means that AO divides `\( \angle A \)` into two equal angles.
Hence, AO bisects `\( \angle A \)`.
In simple words: First, we show that triangle OBC has two equal angles, which means its opposite sides OB and OC are also equal. Then, we prove that triangle AOB and triangle AOC are identical using their sides, which tells us that AO splits angle A exactly in half.

Exam Tip: For geometry proofs, always clearly state your 'Given', 'To Prove', and then provide step-by-step 'Proof' with reasons for each statement. Congruence rules (SSS, SAS, ASA, AAS, RHS) are key tools.

Question 2. In ∆ABC, AD is the perpendicular bisector of BC (see in figure). Show that ΔΑBC is an isosceles triangle in which AB = AC.
Answer:
Given that `\( \triangle ABC \)` has AD as the perpendicular bisector of BC.
This implies two things:
1. AD is perpendicular to BC, so `\( \angle ADB = \angle ADC = 90^\circ \)`.
2. AD bisects BC, so `\( BD = CD \)`.

Now, let's consider `\( \triangle ADB \)` and `\( \triangle ADC \)`.
We have:
1. `\( BD = CD \)` (Because AD is the bisector of BC, as given).
2. `\( \angle ADB = \angle ADC \)` (Both are `\( 90^\circ \)` because AD is perpendicular to BC).
3. `\( AD = AD \)` (This is a common side to both triangles).

By the SAS (Side-Angle-Side) congruence rule, `\( \triangle ADB \cong \triangle ADC \)`.
Since these triangles are congruent, their corresponding parts must be equal (CPCT).
Therefore, `\( AB = AC \)`.
Since two sides of `\( \triangle ABC \)` are equal (AB and AC), `\( \triangle ABC \)` is an isosceles triangle.
In simple words: Because AD cuts BC in half and forms a right angle, we can show that the two smaller triangles (ADB and ADC) are identical. This means their matching sides, AB and AC, must be the same length, making the big triangle ABC an isosceles triangle.

Exam Tip: Remember the definition of a perpendicular bisector: it both cuts a segment into two equal parts and forms a 90-degree angle with it. This creates conditions for triangle congruence, which is often used to prove properties like isosceles triangles.

Question 3. ABC is an isosceles triangle in which altitudes BF and CF are drawn to equal sides AC and AB respectively (See figure). Show that these altitudes are equal.
Answer:
Given that `\( \triangle ABC \)` is an isosceles triangle with `\( AB = AC \)`.
Also, BF is an altitude to side AC, so `\( BF \perp AC \implies \angle BFA = 90^\circ \)`.
And CF is an altitude to side AB, so `\( CF \perp AB \implies \angle CFA = 90^\circ \)`.

Now, let's consider `\( \triangle ABF \)` and `\( \triangle ACF \)`.
We have:
1. `\( \angle AFB = \angle AFC \)` (Both are `\( 90^\circ \)` as BF and CF are altitudes). *Correction: The problem states BE and CF. The OCR solution uses ABE and ACF. I will follow the OCR solution which matches the diagram, despite the question text error.* Let's use `\( \triangle ABE \)` and `\( \triangle ACF \)` as the OCR suggests and the image shows. The image actually shows BE and CF, and the solution refers to them as `\( \triangle ABE \)` and `\( \triangle ACF \)`. So the question has a typo, and the diagram/solution correct it. Let's stick to `BE` and `CF` being altitudes. Let's retry this proof carefully matching the question text, diagram, and solution's intent. The question says "altitudes BF and CF are drawn to equal sides AC and AB respectively". This implies altitude from B is BF to AC, and altitude from C is CF to AB. The diagram shows altitude from B goes to E on AC. And altitude from C goes to F on AB. The solution uses `\( \triangle ABE \)` and `\( \triangle ACF \)`. This implies the altitudes are BE and CF. I will use BE and CF as the solution and diagram imply. Given that `\( \triangle ABC \)` is an isosceles triangle with `\( AB = AC \)`.
BE is an altitude to side AC, so `\( BE \perp AC \implies \angle BEA = 90^\circ \)`.
CF is an altitude to side AB, so `\( CF \perp AB \implies \angle CFA = 90^\circ \)`.

Now, consider `\( \triangle ABE \)` and `\( \triangle ACF \)`.
We have:
1. `\( \angle AEB = \angle AFC \)` (Both are `\( 90^\circ \)` because BE and CF are altitudes).
2. `\( \angle BAE = \angle CAF \)` (This is the common angle `\( \angle A \)` for both triangles).
3. `\( AB = AC \)` (Given, as `\( \triangle ABC \)` is an isosceles triangle).

By the AAS (Angle-Angle-Side) congruence rule, `\( \triangle ABE \cong \triangle ACF \)`.
Since these triangles are congruent, their corresponding parts must be equal (CPCT).
Therefore, `\( BE = CF \)`.
Hence, the altitudes BE and CF are equal.
In simple words: Since the main triangle ABC has two equal sides, and we have two heights (altitudes) drawn from the base angles to those sides, we can show that the two smaller triangles formed (ABE and ACF) are identical. This makes the two heights themselves equal in length.

Exam Tip: When proving altitudes or medians are equal, look for pairs of triangles that share a common angle and have one of the given equal sides. AAS congruence is very useful in these cases.

Question 4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that
(i) ΔΑΒΕ = ΔΑCF
(ii) AB = AC (i.e. ABC is an isosceles triangle
Answer:
Given that BE and CF are altitudes to sides AC and AB respectively, and `\( BE = CF \)`.
This implies:
`\( \angle AEB = 90^\circ \)` (BE is altitude to AC)
`\( \angle AFC = 90^\circ \)` (CF is altitude to AB)

(i) **To show `\( \triangle ABE \cong \triangle ACF \)`:**
Consider `\( \triangle ABE \)` and `\( \triangle ACF \)`.
1. `\( \angle AEB = \angle AFC \)` (Both are `\( 90^\circ \)`, as shown above).
2. `\( \angle BAE = \angle CAF \)` (This is the common angle `\( \angle A \)` for both triangles).
3. `\( BE = CF \)` (Given that the altitudes are equal).

By the AAS (Angle-Angle-Side) congruence rule, `\( \triangle ABE \cong \triangle ACF \)`.

(ii) **To show `\( AB = AC \)`:**
Since `\( \triangle ABE \cong \triangle ACF \)` (proved in part i), their corresponding parts are equal (CPCT).
Therefore, `\( AB = AC \)`.
Because two sides of `\( \triangle ABC \)` are equal (AB and AC), `\( \triangle ABC \)` is an isosceles triangle.
In simple words: First, we prove that the two smaller triangles (ABE and ACF) are identical because they share an angle, both have a right angle, and their heights are equal. Since these triangles are the same, their matching sides AB and AC must be equal, which makes the main triangle ABC an isosceles triangle.

Exam Tip: When proving an isosceles triangle, showing two sides are equal or two angles are equal is sufficient. AAS congruence is a powerful tool when you have two angles and a non-included side.

Question 5. ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that ∠ABD = ∠ACD.
Answer:
Given that `\( \triangle ABC \)` is an isosceles triangle with base BC.
Therefore, `\( AB = AC \)`.
In `\( \triangle ABC \)`, the angles opposite to the equal sides are equal.
So, `\( \angle ABC = \angle ACB \)` ........(1)

Also given that `\( \triangle DBC \)` is an isosceles triangle with base BC.
Therefore, `\( BD = CD \)`.
In `\( \triangle DBC \)`, the angles opposite to the equal sides are equal.
So, `\( \angle DBC = \angle DCB \)` ........(2)

Now, let's add equation (1) and equation (2):
`\( \angle ABC + \angle DBC = \angle ACB + \angle DCB \)`
From the figure, we can see that:
`\( \angle ABC + \angle DBC = \angle ABD \)`
And `\( \angle ACB + \angle DCB = \angle ACD \)`
Therefore, substituting these into the combined equation, we get:
`\( \angle ABD = \angle ACD \)`.
In simple words: Since both triangles ABC and DBC are isosceles and share the same bottom side, their base angles are equal. When we add the corresponding base angles from both triangles, the total angle on the left side becomes equal to the total angle on the right side.

Exam Tip: When dealing with two triangles sharing a base, always remember that angles opposite equal sides are equal. This property is crucial for combining angles to prove larger angle equalities.

Question 6. ∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. (see figure). Show that ∠BCD is a right angle.
Answer:
Given that `\( \triangle ABC \)` is an isosceles triangle with `\( AB = AC \)`.
In `\( \triangle ABC \)`, angles opposite to equal sides are equal.
So, `\( \angle ABC = \angle ACB \)` ........(1)

Side BA is produced to D such that `\( AD = AB \)`.
Since `\( AB = AC \)` (given) and `\( AD = AB \)` (by construction), it follows that `\( AD = AC \)`.
Now, consider `\( \triangle ACD \)`. Since `\( AD = AC \)`, `\( \triangle ACD \)` is an isosceles triangle.
In `\( \triangle ACD \)`, angles opposite to equal sides are equal.
So, `\( \angle ADC = \angle ACD \)` ........(2)

Now, let's add equation (1) and equation (2):
`\( \angle ABC + \angle ADC = \angle ACB + \angle ACD \)`
From the figure, we can see that `\( \angle ACB + \angle ACD = \angle BCD \)`.
So, `\( \angle ABC + \angle ADC = \angle BCD \)` ........(3)

Now, consider the large triangle `\( \triangle BCD \)`. The sum of angles in any triangle is `\( 180^\circ \)`.
`\( \angle DBC + \angle BDC + \angle BCD = 180^\circ \)`
Here, `\( \angle DBC \)` is the same as `\( \angle ABC \)`.
And `\( \angle BDC \)` is the same as `\( \angle ADC \)`.
So, we can write the equation as: `\( \angle ABC + \angle ADC + \angle BCD = 180^\circ \)`.
From equation (3), we know that `\( \angle ABC + \angle ADC = \angle BCD \)`.
Substitute this into the angle sum equation:
`\( \angle BCD + \angle BCD = 180^\circ \)`
`\( 2 \angle BCD = 180^\circ \)`
`\( \angle BCD = \frac{180^\circ}{2} \)`
`\( \angle BCD = 90^\circ \)`.
Hence, `\( \angle BCD \)` is a right angle.
In simple words: First, we use the fact that ABC is isosceles to say two angles are equal. Then, we extend one side to create another isosceles triangle ACD, which means two more angles are equal. By combining these angle equalities and using the total angle sum in triangle BCD, we prove that the angle BCD must be 90 degrees.

Exam Tip: This problem often involves identifying multiple isosceles triangles and using their properties (angles opposite equal sides are equal) in conjunction with the angle sum property of a large triangle.

Question 7. ABC is a right-angled triangle in which ∠A = 90°, and AB = AC. Find ∠B and ∠C.
Answer:
Given that `\( \triangle ABC \)` is a right-angled triangle with `\( \angle A = 90^\circ \)`.
Also, we are given that `\( AB = AC \)`.

In `\( \triangle ABC \)`, since `\( AB = AC \)`, the angles opposite to these equal sides must also be equal.
So, `\( \angle B = \angle C \)`.

The sum of all angles in any triangle is `\( 180^\circ \)`.
Thus, `\( \angle A + \angle B + \angle C = 180^\circ \)`.
Substitute the known values:
`\( 90^\circ + \angle C + \angle C = 180^\circ \)` (Since `\( \angle B = \angle C \)`)
`\( 90^\circ + 2 \angle C = 180^\circ \)`
Subtract `\( 90^\circ \)` from both sides:
`\( 2 \angle C = 180^\circ - 90^\circ \)`
`\( 2 \angle C = 90^\circ \)`
Divide by 2:
`\( \angle C = \frac{90^\circ}{2} \)`
`\( \angle C = 45^\circ \)`.

Since `\( \angle B = \angle C \)`, then `\( \angle B = 45^\circ \)`.
Therefore, `\( \angle B = 45^\circ \)` and `\( \angle C = 45^\circ \)`.
In simple words: Since the triangle has a right angle and two equal sides, the other two angles must also be equal. Because all angles in a triangle add up to 180 degrees, and one angle is 90 degrees, the remaining 90 degrees must be split evenly between the two equal angles, making each 45 degrees.

Exam Tip: For isosceles right-angled triangles, the two non-right angles are always equal to 45 degrees each. This is a common property to remember for quick calculations.

Question 8. Show that the angles of an equilateral triangle are 60° each.
Answer:
Given that `\( \triangle ABC \)` is an equilateral triangle.
By definition, an equilateral triangle has all three sides equal in length.
So, `\( AB = BC = AC \)`.

Since `\( AB = AC \)`, the angles opposite these sides are equal:
`\( \angle C = \angle B \)` ........(1)

Since `\( BC = AC \)`, the angles opposite these sides are equal:
`\( \angle A = \angle B \)` ........(2)

From equations (1) and (2), we can conclude that all three angles are equal:
`\( \angle A = \angle B = \angle C \)`.
Let `\( x \)` represent the measure of each angle, so `\( \angle A = \angle B = \angle C = x \)`.

The sum of all angles in any triangle is `\( 180^\circ \)`.
So, `\( \angle A + \angle B + \angle C = 180^\circ \)`.
Substitute `\( x \)` for each angle:
`\( x + x + x = 180^\circ \)`
`\( 3x = 180^\circ \)`
Divide by 3:
`\( x = \frac{180^\circ}{3} \)`
`\( x = 60^\circ \)`.

Therefore, `\( \angle A = \angle B = \angle C = 60^\circ \)`.
This shows that each angle of an equilateral triangle measures `\( 60^\circ \)`.
In simple words: An equilateral triangle has all its sides the same length. Because of this, all its angles must also be equal. Since the total degrees in any triangle is 180, and there are three equal angles, each angle must be 180 divided by 3, which is 60 degrees.

Exam Tip: Remember that an equilateral triangle is also equiangular, meaning all its angles are equal. This is a fundamental property to recall for many geometry problems.