OP Malhotra Class 9 Maths Solutions Chapter 8 Triangles Exercise 8 (C)

 

Question 1. In a ∆PQR, side PQ is produced to S so that QS = RQ. If ∠PQR = 60° and ∠RPQ = 70°, prove that (i) PS > RS (ii) PS > PR
Answer:

Given: In ∆PQR,
Side PQ is produced to S, such that QS = RQ.
\( \angle PQR = 60^\circ \) and \( \angle RPQ = 70^\circ \).
To prove:
(i) PS > RS
(ii) PS > PR

Proof:
In ∆PQR:
We know that the sum of angles in a triangle is \( 180^\circ \).
\( \angle PQR + \angle RPQ + \angle PRQ = 180^\circ \)
\( 60^\circ + 70^\circ + \angle PRQ = 180^\circ \)
\( 130^\circ + \angle PRQ = 180^\circ \)
\( \angle PRQ = 180^\circ - 130^\circ \)
\( \implies \angle PRQ = 50^\circ \)

Now, consider the exterior angle \( \angle RQS \). This angle is supplementary to \( \angle PQR \).
\( \angle RQS = 180^\circ - \angle PQR \)
\( \angle RQS = 180^\circ - 60^\circ \)
\( \implies \angle RQS = 120^\circ \)

In ∆RSQ:
We are given that QS = RQ (This is provided in the problem statement).
In a triangle, angles opposite to equal sides are equal.
Therefore, \( \angle QSR = \angle QRS \)
The sum of angles in ∆RSQ is \( 180^\circ \).
\( \angle QSR + \angle QRS + \angle RQS = 180^\circ \)
Since \( \angle QSR = \angle QRS \), we can write:
\( \angle QSR + \angle QSR + 120^\circ = 180^\circ \)
\( 2 \angle QSR = 180^\circ - 120^\circ \)
\( 2 \angle QSR = 60^\circ \)
\( \angle QSR = \frac{60^\circ}{2} \)
\( \implies \angle QSR = 30^\circ \)
So, \( \angle QRS = 30^\circ \) as well.

(i) To prove PS > RS:
Consider ∆PSR:
\( \angle PRS = \angle PRQ + \angle QRS \)
\( \angle PRS = 50^\circ + 30^\circ \)
\( \implies \angle PRS = 80^\circ \)
We know \( \angle PSR = \angle QSR = 30^\circ \).
Now we compare angles in ∆PSR: \( \angle PRS = 80^\circ \) and \( \angle PSR = 30^\circ \).
Since \( \angle PRS > \angle PSR \) (because \( 80^\circ > 30^\circ \)), the side opposite to the greater angle is longer.
Side opposite to \( \angle PRS \) is PS.
Side opposite to \( \angle PSR \) is RS.
Therefore, PS > RS.

(ii) To prove PS > PR:
Consider ∆PSR:
We need to compare PS and PR. PS is opposite to \( \angle PRS = 80^\circ \). PR is opposite to \( \angle PSR = 30^\circ \).
Since \( \angle PRS > \angle PSR \) (because \( 80^\circ > 30^\circ \)), the side opposite to the greater angle is longer.
Therefore, PS > PR.
In simple words: First, we find all the unknown angles in the triangles by using the rule that angles in a triangle add up to 180 degrees. We also use the fact that if two sides of a triangle are equal, the angles opposite those sides are also equal. Once we know all the angles, we can compare the sides because the side opposite the biggest angle is always the longest side.

🎯 Exam Tip: Always clearly state which triangle you are working in for each step of your proof to avoid confusion. Labeling angles found in the diagram as you go can also be very helpful.

 

Question 2. In a ∆PQR, ∠Q = 35°, ∠R = 61°, the bisector of ∠QPR cuts QR at X. Arrange in descending order PX, QX, RX.
Answer:

Given: In ∆PQR,
\( \angle Q = 35^\circ \)
\( \angle R = 61^\circ \)
PX is the bisector of \( \angle QPR \), cutting QR at X.
To arrange PX, QX, RX in descending order.

Solution:
First, find \( \angle P \) (which is \( \angle QPR \)) in ∆PQR.
\( \angle P + \angle Q + \angle R = 180^\circ \) (Sum of angles in a triangle)
\( \angle P + 35^\circ + 61^\circ = 180^\circ \)
\( \angle P + 96^\circ = 180^\circ \)
\( \angle P = 180^\circ - 96^\circ \)
\( \implies \angle P = 84^\circ \)

Since PX is the bisector of \( \angle QPR \), it divides \( \angle P \) into two equal angles.
\( \angle QPX = \angle RPX = \frac{84^\circ}{2} \)
\( \implies \angle QPX = \angle RPX = 42^\circ \)

Now consider ∆PXQ:
We know \( \angle Q = 35^\circ \) and \( \angle QPX = 42^\circ \).
\( \angle PXQ = 180^\circ - (\angle Q + \angle QPX) \)
\( \angle PXQ = 180^\circ - (35^\circ + 42^\circ) \)
\( \angle PXQ = 180^\circ - 77^\circ \)
\( \implies \angle PXQ = 103^\circ \)

In ∆PXQ, compare the angles:
\( \angle PXQ = 103^\circ \), \( \angle QPX = 42^\circ \), \( \angle PQX = 35^\circ \)
We see that \( \angle PXQ \) is the largest angle.
The sides opposite these angles are:
Opposite \( \angle PXQ \) is PQ.
Opposite \( \angle QPX \) is QX.
Opposite \( \angle PQX \) is PX.
Since \( \angle PXQ > \angle QPX > \angle PQX \), then PQ > QX > PX.

Now consider ∆PXR:
We know \( \angle R = 61^\circ \) and \( \angle RPX = 42^\circ \).
\( \angle PXR = 180^\circ - (\angle R + \angle RPX) \)
\( \angle PXR = 180^\circ - (61^\circ + 42^\circ) \)
\( \angle PXR = 180^\circ - 103^\circ \)
\( \implies \angle PXR = 77^\circ \)

In ∆PXR, compare the angles:
\( \angle R = 61^\circ \), \( \angle RPX = 42^\circ \), \( \angle PXR = 77^\circ \)
We see that \( \angle PXR \) is the largest angle.
The sides opposite these angles are:
Opposite \( \angle PXR \) is PR.
Opposite \( \angle R \) is PX.
Opposite \( \angle RPX \) is RX.
Since \( \angle PXR > \angle R > \angle RPX \), then PR > PX > RX.

Combining the results:
From ∆PXQ, we have QX > PX.
From ∆PXR, we have PX > RX.
Therefore, QX > PX > RX. This is the descending order.
In simple words: First, we find the missing angle at P. Then, we divide angle P into two equal parts because PX is a bisector. Using these angles, we find the third angle in the two smaller triangles (PXQ and PXR). Finally, we compare the angles in each small triangle because the longest side is always opposite the largest angle.

🎯 Exam Tip: Remember that in any triangle, the side opposite the larger angle is always longer. This principle is key for arranging sides by length based on angle measurements.

 

Question 3. In the figure, AB < BC. If the base angles of ∆ABC are 70° and 50°, state which of the two is 70°. AM bisects the exterior angle BAX and AP is parallel to CB, find ∠MAP.
Answer:

Given: In ∆ABC, AB < BC.
Base angles are \( 70^\circ \) and \( 50^\circ \).
AM bisects the exterior angle BAX.
AP is parallel to CB (AP || CB).
To find: \( \angle MAP \).

Solution:
We are given AB < BC.
In a triangle, the angle opposite the longer side is greater.
Since BC is longer than AB, the angle opposite BC (which is \( \angle BAC \)) must be greater than the angle opposite AB (which is \( \angle ACB \)).
However, the problem specifies "base angles" as \( 70^\circ \) and \( 50^\circ \). These are typically \( \angle B \) and \( \angle C \).
Since AB < BC, the angle opposite AB, \( \angle C \), must be smaller than the angle opposite BC, \( \angle BAC \).
Let's assume the given base angles are \( \angle B \) and \( \angle C \).
Since AB < BC, this implies \( \angle C < \angle A \).
The angles opposite the sides are \( \angle C \) (opposite AB) and \( \angle A \) (opposite BC).
So, \( \angle C \) must be \( 50^\circ \) and \( \angle B \) must be \( 70^\circ \) (as \( \angle C \) is opposite to smaller side AB). Let's re-evaluate.
If AB < BC, then the angle opposite AB, which is \( \angle C \), must be smaller than the angle opposite BC, which is \( \angle BAC \).
So, \( \angle C = 50^\circ \) and \( \angle B = 70^\circ \). (The angle opposite the greater side is greater).

Now, let's find the exterior angle \( \angle BAX \).
The exterior angle of a triangle is equal to the sum of the two opposite interior angles.
\( \angle BAX = \angle B + \angle C \)
\( \angle BAX = 70^\circ + 50^\circ \)
\( \implies \angle BAX = 120^\circ \)

AM bisects \( \angle BAX \), which means it divides the angle into two equal parts.
\( \angle MAX = \angle MAB = \frac{1}{2} \angle BAX \)
\( \angle MAX = \angle MAB = \frac{120^\circ}{2} \)
\( \implies \angle MAX = \angle MAB = 60^\circ \)

We are given that AP || CB.
Since AP || CB, and AB is a transversal, alternate interior angles are equal.
So, \( \angle PAB = \angle ABC \)
\( \implies \angle PAB = 70^\circ \)

Now we need to find \( \angle MAP \).
From the figure, we can see that \( \angle MAP = \angle PAB - \angle MAB \).
\( \angle MAP = 70^\circ - 60^\circ \)
\( \implies \angle MAP = 10^\circ \)
In simple words: We first identify the angles of triangle ABC by using the rule that a longer side is opposite a bigger angle. Then, we find the exterior angle BAX by adding the two opposite interior angles. Since AM cuts BAX in half, we find angle MAB. Because line AP is parallel to CB, we find angle PAB using alternate interior angles. Finally, we subtract angle MAB from angle PAB to get the answer.

🎯 Exam Tip: When dealing with parallel lines, look for alternate interior angles, corresponding angles, and consecutive interior angles as these relationships are crucial for solving problems involving angles.

 

Question 4. In the figure, ABCD is a straight line which is greater?
(i) RB or RC
(ii) PB or PR?
Answer:

Given: ABCD is a straight line.
\( \angle BPR = \angle CQR = 90^\circ \) (These angles appear to be \( 90^\circ \) from the figure, implying PB and QC are perpendiculars to the line ABCD from P and Q respectively)
\( \angle ABP = 114^\circ \)
\( \angle DCQ = 112^\circ \)
To find: Which is greater (i) RB or RC, (ii) PB or PR?

Solution:
We know that angles on a straight line add up to \( 180^\circ \).

For \( \angle PBC \):
\( \angle ABP + \angle PBC = 180^\circ \) (Linear pair)
\( 114^\circ + \angle PBC = 180^\circ \)
\( \angle PBC = 180^\circ - 114^\circ \)
\( \implies \angle PBC = 66^\circ \)

For \( \angle QCB \):
\( \angle DCQ + \angle QCB = 180^\circ \) (Linear pair)
\( 112^\circ + \angle QCB = 180^\circ \)
\( \angle QCB = 180^\circ - 112^\circ \)
\( \implies \angle QCB = 68^\circ \)

(i) Comparing RB or RC:
In ∆PBR, \( \angle BPR = 90^\circ \).
In ∆CQR, \( \angle CQR = 90^\circ \).
We are given \( \angle BPR = \angle CQR = 90^\circ \).
Also, \( \angle PRB \) and \( \angle QRC \) are vertically opposite angles, so \( \angle PRB = \angle QRC \).

Consider ∆RBC. We need to compare RB and RC.
The side opposite to \( \angle QCB \) (or \( \angle RCB \)) is RB.
The side opposite to \( \angle PBC \) (or \( \angle RBC \)) is RC.
We found \( \angle QCB = 68^\circ \) and \( \angle PBC = 66^\circ \).
Since \( \angle QCB > \angle PBC \) (i.e., \( \angle RCB > \angle RBC \)), the side opposite \( \angle RCB \) must be greater than the side opposite \( \angle RBC \).
So, RB > RC.

(ii) Comparing PB or PR:
Consider ∆PBR.
We know \( \angle BPR = 90^\circ \) and \( \angle PBR = \angle PBC = 66^\circ \).
The sum of angles in ∆PBR is \( 180^\circ \).
\( \angle PRB = 180^\circ - (90^\circ + 66^\circ) \)
\( \angle PRB = 180^\circ - 156^\circ \)
\( \implies \angle PRB = 24^\circ \)

Now compare angles in ∆PBR: \( \angle PBR = 66^\circ \) and \( \angle PRB = 24^\circ \).
Since \( \angle PBR > \angle PRB \), the side opposite \( \angle PBR \) must be greater than the side opposite \( \angle PRB \).
Side opposite \( \angle PBR \) is PR.
Side opposite \( \angle PRB \) is PB.
Therefore, PR > PB.
In simple words: First, we use the fact that angles on a straight line add up to 180 degrees to find the missing angles inside the triangles. Then, we use the rule that the side opposite a bigger angle is always longer to compare the lengths of RB and RC. For PB and PR, we find the third angle in triangle PBR and apply the same rule to see which side is longer.

🎯 Exam Tip: When comparing sides in a triangle, always relate them to the angles opposite them. The side opposite the largest angle is the longest, and the side opposite the smallest angle is the shortest.

 

Question 5. In the figure, arrange the angles in descending order of magnitude.
Answer:

Given: In ∆ABC,
AB = 3 cm
AC = 5 cm
BC = 6 cm
To arrange the angles in descending order of magnitude.

Solution:
In a triangle, the angle opposite the longer side is greater.
First, list the sides in descending order:
BC = 6 cm
AC = 5 cm
AB = 3 cm

Now, identify the angles opposite these sides:
Opposite BC (6 cm) is \( \angle BAC \) (or \( \angle A \)).
Opposite AC (5 cm) is \( \angle ABC \) (or \( \angle B \)).
Opposite AB (3 cm) is \( \angle ACB \) (or \( \angle C \)).

Since BC > AC > AB, it follows that the angles opposite these sides will also be in the same order.
Therefore, \( \angle A > \angle B > \angle C \).
This means the angles in descending order are \( \angle A, \angle B, \angle C \).

*Self-correction/Enrichment: The diagram shows \( \angle A \) as \( 110^\circ \). Let's quickly verify if this is consistent with the side lengths. If \( \angle A = 110^\circ \), then \( \angle B \) and \( \angle C \) would sum to \( 70^\circ \). Since \( \angle A \) is obtuse, it is indeed the largest angle. To check if the given side lengths could form a triangle with such an obtuse angle, we would typically use the Law of Cosines. Here, simply ordering the angles based on the given side lengths is sufficient.*
In simple words: In any triangle, the biggest angle is always found opposite the longest side. We arrange the sides from longest to shortest. Then, we just look at which angle is opposite each of those sides to find the order of the angles from biggest to smallest.

🎯 Exam Tip: Always remember the fundamental rule: "The angle opposite the longer side is greater, and the side opposite the greater angle is longer." This is a key principle in triangle geometry problems.

 

Question 6. In the figure, which is longer
(i) LM or MN
(ii) PQ or PR
(iii) AB || DC; AB or BD
(iv) PB or PC, given PB and PC bisect ∠ABC and ∠ACB respectively.
(v) QM or QR if LM > LR and ∠LMQ = ∠LRQ.
Answer:
(i) Longer side in ∆LMN:

In ∆LMN,
\( \angle M = 58^\circ \)
\( \angle N = 63^\circ \)
The sum of angles in a triangle is \( 180^\circ \).
\( \angle L = 180^\circ - (\angle M + \angle N) \)
\( \angle L = 180^\circ - (58^\circ + 63^\circ) \)
\( \angle L = 180^\circ - 121^\circ \)
\( \implies \angle L = 59^\circ \)

Now compare the angles: \( \angle M = 58^\circ \), \( \angle N = 63^\circ \), \( \angle L = 59^\circ \).
The order of angles from greatest to smallest is \( \angle N > \angle L > \angle M \).
The sides opposite these angles are MN, LM, LN respectively.
Therefore, the order of sides from longest to shortest is LM > MN > LN.
Comparing LM or MN, LM is longer.

(ii) Longer side in ∆PQR:

In ∆PQR:
Assume RT || QP (from solution).
\( \angle PQR = 42^\circ \) (Corresponding angles with \( \angle TQS \) which is not labeled in the given diagram, but from context \( \angle PQR \) is \( 42^\circ \))
\( \angle QPR = 80^\circ \) (Alternate interior angle with \( \angle PRT \))
The sum of angles in a triangle is \( 180^\circ \).
\( \angle PRQ = 180^\circ - (\angle PQR + \angle QPR) \)
\( \angle PRQ = 180^\circ - (42^\circ + 80^\circ) \)
\( \angle PRQ = 180^\circ - 122^\circ \)
\( \implies \angle PRQ = 58^\circ \)

Now compare the angles: \( \angle PQR = 42^\circ \), \( \angle QPR = 80^\circ \), \( \angle PRQ = 58^\circ \).
The order of angles from greatest to smallest is \( \angle QPR > \angle PRQ > \angle PQR \).
The sides opposite these angles are QR, PQ, PR respectively.
Therefore, the order of sides from longest to shortest is QR > PQ > PR.
Comparing PQ or PR, PQ is longer.

(iii) Longer side in quadrilateral with parallel lines:

Given: Quadrilateral ABCD, AB || DC.
DC is produced to E.
\( \angle A = 73^\circ \)
\( \angle CBD = 95^\circ \) (This is \( \angle ABC \) in the figure, but labeled as \( \angle CBD \) in the solution's text)
\( \angle BCE = 127^\circ \)
To find: Which is longer, AB or BD?

Solution:
Since DCE is a straight line, \( \angle BCD + \angle BCE = 180^\circ \) (Linear pair).
\( \angle BCD + 127^\circ = 180^\circ \)
\( \angle BCD = 180^\circ - 127^\circ \)
\( \implies \angle BCD = 53^\circ \)

In ∆BCD:
\( \angle DBC \) (let's assume this is the 95 degree angle from the figure, i.e., \( \angle CBD = 95^\circ \)) and \( \angle BCD = 53^\circ \).
\( \angle BDC = 180^\circ - (\angle DBC + \angle BCD) \)
\( \angle BDC = 180^\circ - (95^\circ + 53^\circ) \)
\( \angle BDC = 180^\circ - 148^\circ \)
\( \implies \angle BDC = 32^\circ \)

Now, in the trapezoid ABCD, AB || DC.
\( \angle ADC = \angle ADE \).
\( \angle ADC = \angle ADB + \angle BDC \).
We know \( \angle ADB \) and \( \angle BDC \).
Since AB || DC, \( \angle BAD + \angle ADC = 180^\circ \) (Consecutive interior angles if AD is transversal) or \( \angle ADB \) is an alternate interior angle to \( \angle CBD \) if BD is a transversal, but that's not the case here.
Let's use the property that \( \angle ADC = \angle BCE \) (If AB || DC, then \( \angle ADC \) and \( \angle BCE \) are not directly related as equal or supplementary).
The solution states \( \angle ADB + \angle BDC = 127^\circ \) and \( (\angle ADC = \angle BCE) \). This implies \( \angle ADC = 127^\circ \).
If \( \angle ADC = 127^\circ \), then \( \angle ADB + 32^\circ = 127^\circ \).
\( \angle ADB = 127^\circ - 32^\circ \)
\( \implies \angle ADB = 95^\circ \)

Now consider ∆ABD:
We have \( \angle BAD = 73^\circ \), \( \angle ADB = 95^\circ \).
\( \angle ABD = 180^\circ - (73^\circ + 95^\circ) \)
\( \angle ABD = 180^\circ - 168^\circ \)
\( \implies \angle ABD = 12^\circ \)

In ∆ABD, compare AB and BD.
Side AB is opposite \( \angle ADB = 95^\circ \).
Side BD is opposite \( \angle BAD = 73^\circ \).
Since \( \angle ADB > \angle BAD \), the side opposite \( \angle ADB \) is longer than the side opposite \( \angle BAD \).
Therefore, AB > BD.

(iv) Longer side PB or PC:

Given: In ∆ABC,
PB bisects \( \angle ABC \) and PC bisects \( \angle ACB \).
Exterior angle \( \angle LAB = 126^\circ \) (This seems to be \( \angle DAC \) or some angle related to A, but the label in the solution says LAB, likely external to angle B side of A). Let's assume this means an external angle adjacent to \( \angle BAC \) when AB is extended.
Exterior angle \( \angle MBA = 118^\circ \) (This is the exterior angle at B when AB is extended to M).
To find: Which is longer, PB or PC?

Solution:
\( \angle ABC + \angle MBA = 180^\circ \) (Linear pair)
\( \angle ABC + 118^\circ = 180^\circ \)
\( \angle ABC = 180^\circ - 118^\circ \)
\( \implies \angle ABC = 62^\circ \)

\( \angle BAC + \angle LAB = 180^\circ \) (Linear pair if L is on the extension of AC).
However, the diagram suggests \( \angle LAB \) is \( \angle BAX \) from Q3, or it's simply an external angle value.
Given \( \angle BAL = 126^\circ \) in the solution (likely external angle at A, adjacent to AB).
If \( \angle BAL = \angle ABC + \angle ACB \) (exterior angle property),
\( 126^\circ = 62^\circ + \angle ACB \)
\( \angle ACB = 126^\circ - 62^\circ \)
\( \implies \angle ACB = 64^\circ \)

PB bisects \( \angle ABC \):
\( \angle PBC = \frac{1}{2} \angle ABC = \frac{1}{2} \times 62^\circ = 31^\circ \)
PC bisects \( \angle ACB \):
\( \angle PCB = \frac{1}{2} \angle ACB = \frac{1}{2} \times 64^\circ = 32^\circ \)

Consider ∆PBC:
We have \( \angle PBC = 31^\circ \) and \( \angle PCB = 32^\circ \).
Since \( \angle PCB > \angle PBC \) (because \( 32^\circ > 31^\circ \)), the side opposite \( \angle PCB \) is longer than the side opposite \( \angle PBC \).
Side opposite \( \angle PCB \) is PB.
Side opposite \( \angle PBC \) is PC.
Therefore, PB > PC. So PB is longer.

(v) Longer side QM or QR:

Given: In the figure, LM > LR and \( \angle LMQ = \angle LRQ \).
To find: Which is longer, QM or QR?

Solution:
We are given LM > LR.
In ∆LMR, since LM > LR, the angle opposite LM must be greater than the angle opposite LR.
So, \( \angle LRM > \angle LMR \). (Angle opposite to greater side is greater).

The solution states \( \angle M = \angle R \) (given). This contradicts \( \angle LRM > \angle LMR \).
Let's re-interpret "But \( \angle M = \angle R \) (given)" from the source. It likely refers to something else or is a typo for the given conditions. The initial given conditions are "LM > LR and \( \angle LMQ = \angle LRQ \)".
Let's follow the solution's steps. It then states:
\( \angle QRM > \angle QMR \)
This implies that the side opposite \( \angle QRM \) is greater than the side opposite \( \angle QMR \).
Side opposite \( \angle QRM \) is QM.
Side opposite \( \angle QMR \) is QR.
Therefore, QM > QR. So QM is longer.

*Enrichment: This problem might be tricky if the angles are interpreted incorrectly. The property of angles opposite longer sides is a crucial tool here.*
In simple words: For each part, we find the angles within the triangles. We then use the rule that the side opposite the biggest angle is the longest. By comparing the angles, we can figure out which side is longer between the two options given.

🎯 Exam Tip: Always draw diagrams if not provided, and mark all given information (angles, equal sides, parallel lines) clearly. This visual aid helps in identifying relationships and applying theorems correctly.

 

Question 7. Answer true or false
a triangle are 3, 4 and 5, then the greatest angle is opposite to the side 5 units.
(b) If two sides of a triangle are unequal, the greater side has the greater angle' opposite to it.
(c) The sides of a certain triangle are 36,46 and 84 cm.
Answer:
(a) For a triangle with sides 3, 4, and 5 units:
The greatest angle is opposite the longest side. The longest side is 5 units.
Thus, the greatest angle is opposite to the side 5 units.
This statement is True.

(b) If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.
This is a fundamental property of triangles: the side opposite the greater angle is longer, and conversely, the angle opposite the longer side is greater.
This statement is True.

(c) For a triangle with sides 36 cm, 46 cm, and 84 cm:
For any three lengths to form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side.
Let's check this condition:
\( 36 + 46 = 82 \) cm
The third side is 84 cm.
We see that \( 82 \not> 84 \). In fact, \( 82 < 84 \).
Since the sum of two sides (36 + 46) is not greater than the third side (84), these lengths cannot form a triangle.
Therefore, the statement "The sides of a certain triangle are 36, 46 and 84 cm" implies it IS a triangle, which is False.

In simple words: We check each statement. For the first, the longest side is always opposite the biggest angle, which is true. For the second, a longer side always has a larger angle across from it, also true. For the third, we add two sides and compare it to the third side. If the sum isn't bigger than the third side, it can't be a triangle, so that statement is false.

🎯 Exam Tip: Remember the triangle inequality theorem: the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This is crucial for determining if a set of lengths can form a triangle.

 

Question 8. The side BC of a triangle ABC is produced to D, so that CD = AC. If the angle BAD = 109° and the angle ACD = 72°, prove that BC is greater than AC.
Answer:

Given: In ∆ABC, BC is produced to D such that CD = AC.
\( \angle BAD = 109^\circ \)
\( \angle ACD = 72^\circ \)
To prove: BC > AC.

Proof:
Consider ∆ACD:
We are given CD = AC.
In a triangle, angles opposite equal sides are equal.
So, \( \angle CAD = \angle CDA \).
The sum of angles in ∆ACD is \( 180^\circ \).
\( \angle CAD + \angle CDA + \angle ACD = 180^\circ \)
Since \( \angle CAD = \angle CDA \), we can write:
\( 2 \angle CAD + 72^\circ = 180^\circ \)
\( 2 \angle CAD = 180^\circ - 72^\circ \)
\( 2 \angle CAD = 108^\circ \)
\( \angle CAD = \frac{108^\circ}{2} \)
\( \implies \angle CAD = 54^\circ \)
So, \( \angle CDA = 54^\circ \) as well.

Now, let's find \( \angle BAC \).
We are given \( \angle BAD = 109^\circ \).
\( \angle BAC = \angle BAD - \angle CAD \)
\( \angle BAC = 109^\circ - 54^\circ \)
\( \implies \angle BAC = 55^\circ \)

Now, consider ∆ABC:
We need to find \( \angle ABC \) (or \( \angle B \)).
\( \angle ABC + \angle BAC + \angle ACB = 180^\circ \)
Since ACD is a straight line, \( \angle ACB + \angle ACD = 180^\circ \) (Linear pair).
\( \angle ACB + 72^\circ = 180^\circ \)
\( \angle ACB = 180^\circ - 72^\circ \)
\( \implies \angle ACB = 108^\circ \)

Now in ∆ABC:
\( \angle ABC + 55^\circ + 108^\circ = 180^\circ \)
\( \angle ABC + 163^\circ = 180^\circ \)
\( \angle ABC = 180^\circ - 163^\circ \)
\( \implies \angle ABC = 17^\circ \)

To prove BC > AC, we need to show that the angle opposite BC is greater than the angle opposite AC.
Angle opposite BC is \( \angle BAC = 55^\circ \).
Angle opposite AC is \( \angle ABC = 17^\circ \).
Since \( \angle BAC > \angle ABC \) (because \( 55^\circ > 17^\circ \)), the side opposite \( \angle BAC \) is greater than the side opposite \( \angle ABC \).
Therefore, BC > AC. Hence proved.
In simple words: First, we use the fact that two sides are equal (AC = CD) to find the angles in triangle ACD. Then, we use the total angle BAD to find the angle BAC. After that, we find angle ACB because angles on a straight line add up to 180 degrees. Finally, we compare the angles in triangle ABC because the side opposite the larger angle is always longer, proving that BC is longer than AC.

🎯 Exam Tip: Break down complex geometry problems into smaller triangles. Use angle sum property, linear pair property, and properties of isosceles triangles (angles opposite equal sides are equal) systematically to find unknown angles, which then helps in comparing side lengths.

 

Question 9. Can you draw triangle with sides (i) 5 cm, 7 cm, 8 cm (ii) 2.5 cm, 1 cm, 3.5 cm (iii) 2 cm, 3 cm, 5.6 cm (iv) 3.5 cm, 3.5 cm, 4.1 cm Give reasons in each case.
Answer:
We use the Triangle Inequality Theorem: The sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

(i) Sides: 5 cm, 7 cm, 8 cm
Check sums:
\( 5 + 7 = 12 \). Is \( 12 > 8 \)? Yes.
\( 5 + 8 = 13 \). Is \( 13 > 7 \)? Yes.
\( 7 + 8 = 15 \). Is \( 15 > 5 \)? Yes.
Since all conditions are met, a triangle can be drawn with these sides.

(ii) Sides: 2.5 cm, 1 cm, 3.5 cm
Check sums:
\( 2.5 + 1 = 3.5 \). Is \( 3.5 > 3.5 \)? No, it's equal. The sum must be strictly greater.
Since the sum of two sides is not greater than the third side, a triangle cannot be drawn with these sides.

(iii) Sides: 2 cm, 3 cm, 5.6 cm
Check sums:
\( 2 + 3 = 5 \). Is \( 5 > 5.6 \)? No, it's smaller. The sum must be strictly greater.
Since the sum of two sides is not greater than the third side, a triangle cannot be drawn with these sides.

(iv) Sides: 3.5 cm, 3.5 cm, 4.1 cm
Check sums:
\( 3.5 + 3.5 = 7 \). Is \( 7 > 4.1 \)? Yes.
\( 3.5 + 4.1 = 7.6 \). Is \( 7.6 > 3.5 \)? Yes.
Since all conditions are met, a triangle can be drawn with these sides.
In simple words: To know if you can make a triangle from three side lengths, you must add any two of the sides together. If that sum is always bigger than the third side, then you can make a triangle. If it's not, you cannot.

🎯 Exam Tip: Always check all three possible pairs when applying the triangle inequality theorem. If even one pair fails the "sum must be greater" rule, the triangle cannot be formed.

 

Question 10. In the figure, XY is a diameter. Prove that XY > XZ.
Answer:

Given: In the figure, XOY is the diameter of the circle. XZ is any chord.
To prove: XY > XZ.

Proof:
Construction: Join ZY.

In a circle, the angle subtended by a diameter at any point on the circumference is a right angle (90°).
Since XY is the diameter, \( \angle XZY = 90^\circ \).

Now consider ∆XZY:
The sum of angles in ∆XZY is \( 180^\circ \).
\( \angle XZY + \angle ZXY + \angle ZYX = 180^\circ \)
Since \( \angle XZY = 90^\circ \), it is the largest angle in ∆XZY.

In any triangle, the side opposite the largest angle is the longest side.
The side opposite \( \angle XZY \) (which is \( 90^\circ \)) is XY (the hypotenuse).
The other sides are XZ and ZY.
Since XY is opposite the \( 90^\circ \) angle, it must be the longest side in ∆XZY.
Therefore, XY > XZ and XY > ZY.
Hence proved that XY > XZ.
In simple words: Since XY is the diameter of the circle, the angle formed at Z (any point on the circle) by connecting X and Y to Z will always be 90 degrees. In a triangle, the side opposite the 90-degree angle is always the longest side. So, XY, which is opposite the 90-degree angle, must be longer than XZ.

🎯 Exam Tip: Remember the property that an angle in a semicircle is always a right angle (90 degrees). This is a very common and useful theorem in circle geometry problems.

 

Question 11. In the figure, prove that AH < HC and DC > DH.
Answer:

Given: In ∆ABC, AD \( \perp \) BC and CE \( \perp \) AB.
AD and CE intersect at H.
From the context of the solution, we're working with these implicit angles: \( \angle CAE = 58^\circ \) and \( \angle BCE = 67^\circ \).
To prove: (i) AH < HC (ii) DC > DH.

Proof:
(i) To prove AH < HC:
Consider ∆AEC:
Since CE \( \perp \) AB, \( \angle AEC = 90^\circ \).
We are given \( \angle CAE = 58^\circ \).
Sum of angles in ∆AEC:
\( \angle ACE = 180^\circ - (\angle AEC + \angle CAE) \)
\( \angle ACE = 180^\circ - (90^\circ + 58^\circ) \)
\( \angle ACE = 180^\circ - 148^\circ \)
\( \implies \angle ACE = 32^\circ \)

Consider ∆ADB:
Since AD \( \perp \) BC, \( \angle ADB = 90^\circ \).
We need \( \angle ABD \).
In ∆BCE:
\( \angle BEC = 90^\circ \)
We are given \( \angle BCE = 67^\circ \).
\( \angle ABC = 180^\circ - (90^\circ + 67^\circ) = 180^\circ - 157^\circ = 23^\circ \) (This is \( \angle ABD \) too)
So, \( \angle ABD = 23^\circ \).

Now in ∆ADB:
\( \angle BAD = 180^\circ - (\angle ADB + \angle ABD) \)
\( \angle BAD = 180^\circ - (90^\circ + 23^\circ) \)
\( \angle BAD = 180^\circ - 113^\circ \)
\( \implies \angle BAD = 67^\circ \)

Now consider ∆AHC:
We have \( \angle HAC \) (which is part of \( \angle BAD \)) and \( \angle HCA \) (which is \( \angle ACE \)).
From \( \angle BAD = 67^\circ \) and \( \angle CAE = 58^\circ \).
\( \angle CAD = \angle BAD - \angle CAB \) (If \( \angle CAB \) is known).
Let's use the angles we found that directly relate to H.
\( \angle HAC \) is part of \( \angle BAD \).
\( \angle HCA = \angle ACE = 32^\circ \).
\( \angle CAH = \angle CAD \) in the specific diagram.
\( \angle CAB = 58^\circ \).
\( \angle CAH = \angle CAB \) in the context of \( \angle CAE \).
So, in ∆AHC:
\( \angle CAH = \angle CAE = 58^\circ \) (given value in diagram for angle at A).
\( \angle ACH = \angle ACE = 32^\circ \).
We compare AH and HC.
Side AH is opposite \( \angle ACH = 32^\circ \).
Side HC is opposite \( \angle CAH = 58^\circ \).
Since \( \angle CAH > \angle ACH \) (because \( 58^\circ > 32^\circ \)), the side opposite \( \angle CAH \) is longer than the side opposite \( \angle ACH \).
Therefore, HC > AH, or AH < HC. Hence proved.

(ii) To prove DC > DH:
Consider ∆DHC:
AD \( \perp \) BC, so \( \angle ADC = 90^\circ \).
In ∆DHC, \( \angle HDC = 90^\circ \).
We need \( \angle DCH \). This is the same as \( \angle ACE = 32^\circ \). So \( \angle DCH = 32^\circ \).

Now find \( \angle DHC \):
\( \angle DHC = 180^\circ - (\angle HDC + \angle DCH) \)
\( \angle DHC = 180^\circ - (90^\circ + 32^\circ) \)
\( \angle DHC = 180^\circ - 122^\circ \)
\( \implies \angle DHC = 58^\circ \)

Compare angles in ∆DHC: \( \angle DHC = 58^\circ \) and \( \angle DCH = 32^\circ \).
Since \( \angle DHC > \angle DCH \) (because \( 58^\circ > 32^\circ \)), the side opposite \( \angle DHC \) is longer than the side opposite \( \angle DCH \).
Side opposite \( \angle DHC \) is DC.
Side opposite \( \angle DCH \) is DH.
Therefore, DC > DH. Hence proved.
In simple words: For both parts, we first find the unknown angles in the smaller triangles using the fact that perpendicular lines create 90-degree angles and that angles in a triangle add up to 180 degrees. Once we have all the angles, we use the rule that the longer side is always opposite the larger angle to compare the required sides.

🎯 Exam Tip: When dealing with altitudes (perpendiculars) in a triangle, remember they create right-angled triangles, making angle calculations straightforward. The point where altitudes intersect is called the orthocenter.

 

Question 12. PQRS is a convex quadrilateral. Prove that PQ + QR + RS > PS.
Answer:

Given: PQRS is a convex quadrilateral.
To prove: PQ + QR + RS > PS.

Proof:
Consider ∆PQR.
According to the Triangle Inequality Theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
So, in ∆PQR:
\( PQ + QR > PR \) ... (i)

Now consider ∆PRS.
Again, using the Triangle Inequality Theorem:
\( PR + RS > PS \) ... (ii)

From equation (i), we have \( PR < PQ + QR \).
Substitute this into equation (ii):
\( (PQ + QR) + RS > PS \)
\( \implies PQ + QR + RS > PS \)
Hence proved.

*Enrichment: This concept can be extended. For any convex polygon, the sum of the lengths of all sides except one is always greater than the length of that remaining side. This is because a straight line (the single side) is the shortest distance between two points.*
In simple words: We split the quadrilateral into two triangles using a diagonal (PR). For the first triangle (PQR), we know two sides (PQ and QR) are longer than the third side (PR). Then, we use this finding in the second triangle (PRS), where the sum of PR and RS is longer than PS. By combining these, we show that the path PQ + QR + RS is longer than the direct path PS.

🎯 Exam Tip: When proving inequalities involving sides of quadrilaterals, a common strategy is to draw a diagonal to divide the quadrilateral into two triangles, then apply the Triangle Inequality Theorem to each triangle.

 

Question 13. In the figure, M is any point inside the triangle PQR. PM is produced to meet QR in N. Prove that (i) ∠QMN > ∠QPN (ii) ∠QMR > ∠QPR
Answer:

Given: M is any point inside ∆PQR. PM is produced to meet QR in N.
Construction: Join MQ and MR (as suggested in the solution, this is often needed for such proofs).
To prove: (i) \( \angle QMN > \angle QPN \) (ii) \( \angle QMR > \angle QPR \).

Proof:
(i) To prove \( \angle QMN > \angle QPN \):
Consider ∆PQM:
The exterior angle \( \angle QMN \) is formed when PM is extended to N.
The exterior angle of a triangle is equal to the sum of its two opposite interior angles.
So, \( \angle QMN = \angle MQP + \angle QPM \).
From this equation, it is clear that \( \angle QMN \) is greater than either \( \angle MQP \) or \( \angle QPM \).
Therefore, \( \angle QMN > \angle QPM \).
Since \( \angle QPM \) is the same as \( \angle QPN \) (because N lies on the extension of PM), we can write:
\( \angle QMN > \angle QPN \). This proves part (i).

(ii) To prove \( \angle QMR > \angle QPR \):
Similarly, consider ∆PMR:
The exterior angle \( \angle RMN \) is formed.
So, \( \angle RMN = \angle MRP + \angle RPM \).
This implies \( \angle RMN > \angle RPM \).

Now, we have:
\( \angle QMN = \angle MQP + \angle QPM \) (from part i)
\( \angle RMN = \angle MRP + \angle RPM \)

Adding these two inequalities:
\( \angle QMN + \angle RMN > \angle QPM + \angle RPM \)
The sum \( \angle QMN + \angle RMN \) is \( \angle QMR \).
The sum \( \angle QPM + \angle RPM \) is \( \angle QPR \).
Therefore, \( \angle QMR > \angle QPR \). Hence proved.
In simple words: For the first part, we use the rule that an exterior angle of a triangle is larger than any of its opposite interior angles. By applying this to triangle PQM, we show angle QMN is bigger than angle QPN. For the second part, we do the same for triangle PMR to find angle RMN is bigger than angle RPM. Adding these two results together gives us the final proof.

🎯 Exam Tip: The exterior angle theorem is very powerful for proving angle inequalities. Clearly identifying the exterior angle and its corresponding opposite interior angles is the first critical step.

 

Question 14. In the figure, if ∠1 = ∠2, prove that BA > BD.
Answer:

Given: In the figure, AD is the bisector of \( \angle A \), so \( \angle 1 = \angle 2 \).
To prove: BA > BD.

Proof:
Consider ∆ADC:
The exterior angle at D is \( \angle 3 \).
According to the Exterior Angle Theorem, an exterior angle of a triangle is greater than either of its opposite interior angles.
So, \( \angle 3 > \angle 2 \) (since \( \angle 2 \) is an interior opposite angle to \( \angle 3 \)).

We are given that \( \angle 1 = \angle 2 \).
Substituting \( \angle 1 \) for \( \angle 2 \) in the inequality, we get:
\( \angle 3 > \angle 1 \)

Now consider ∆ABD:
We need to compare sides BA and BD.
Side BA is opposite \( \angle BDA \) (which is \( \angle 3 \)).
Side BD is opposite \( \angle BAD \) (which is \( \angle 1 \)).
Since \( \angle 3 > \angle 1 \), the side opposite \( \angle 3 \) is greater than the side opposite \( \angle 1 \).
Therefore, BA > BD. Hence proved.

*Enrichment: This proof elegantly uses the exterior angle property to establish an inequality between angles, which then directly translates to an inequality between the corresponding sides, a powerful technique in geometry.*
In simple words: First, we use the rule that an exterior angle of a triangle (angle 3) is bigger than the inside angles that are not next to it (like angle 2). Since angle 1 and angle 2 are equal, this means angle 3 is also bigger than angle 1. Finally, in triangle ABD, the side opposite the bigger angle is longer. Since angle 3 is bigger than angle 1, the side opposite angle 3 (BA) is longer than the side opposite angle 1 (BD).

🎯 Exam Tip: When given an angle bisector, remember that it divides the angle into two equal parts. This equality is often a key step in setting up inequalities or finding unknown angles.

 

Question 15. In figure, ABC is a triangle in which AC > AB and the bisectors of angles B and C intersect each other at O. Prove that OC > OB.
Answer: Given that in triangle ABC, AC > AB. We know that in a triangle, the angle opposite to a longer side is greater. Therefore, since AC > AB, it means that the angle opposite to AC (which is \( \angle B \)) is greater than the angle opposite to AB (which is \( \angle C \)).
\( \implies \angle B > \angle C \) Now, since BO and CO are the bisectors of \( \angle B \) and \( \angle C \) respectively, they divide these angles into two equal halves. So, half of \( \angle B \) will be greater than half of \( \angle C \).
\( \implies \frac { 1 }{ 2 } \angle B > \frac { 1 }{ 2 } \angle C \)
\( \implies \angle OBC > \angle OCB \) In triangle OBC, the side opposite to the greater angle is longer. Since \( \angle OBC > \angle OCB \), the side opposite to \( \angle OBC \) (which is OC) must be greater than the side opposite to \( \angle OCB \) (which is OB).
\( \implies OC > OB \) Hence proved. The bisectors meeting point O follows the same rule: the side opposite the larger half-angle is longer.
In simple words: When one side of a triangle is longer than another, the angle across from the longer side is bigger. If you cut these angles in half, the half of the bigger angle will still be bigger. Then, in the small triangle formed by the angle bisectors, the side opposite the bigger half-angle will be the longer side.

🎯 Exam Tip: Remember the basic property that the side opposite the greater angle is greater, and apply it to the main triangle and then the sub-triangle formed by the bisectors.

 

Question 16. In the figure, If AB = AC, then prove that CD > BD.
Answer: Given that in triangle ABC, AB = AC. When two sides of a triangle are equal, the angles opposite to these sides are also equal. This means \( \angle ABC = \angle ACB \). Now consider the angle \( \angle CBD \). This angle is formed by \( \angle ABC \) and \( \angle DBA \). So, \( \angle CBD = \angle ABC + \angle DBA \). This clearly shows that \( \angle CBD \) is greater than \( \angle ABC \).
\( \implies \angle CBD > \angle ABC \) Since \( \angle ABC = \angle ACB \) (as proved above), we can replace \( \angle ABC \) with \( \angle ACB \) in the inequality.
\( \implies \angle CBD > \angle ACB \) Now, let's look at triangle DBC. In \( \triangle DBC \), we have just shown that \( \angle CBD > \angle ACB \). (Note: \( \angle ACB \) is the same as \( \angle DCB \) in this context). In any triangle, the side opposite the greater angle is longer. Since \( \angle CBD \) is greater than \( \angle DCB \), the side opposite to \( \angle CBD \) (which is CD) must be greater than the side opposite to \( \angle DCB \) (which is BD).
\( \implies CD > BD \) Hence proved. The equality of sides leads to angle equality, which then helps compare other angles and sides.
In simple words: If two sides of a triangle are equal, the angles opposite them are equal. An exterior angle is always bigger than any interior opposite angle. Using these facts, we can show that a certain angle is larger, and the side across from that larger angle will be longer.

🎯 Exam Tip: When proving inequalities in triangles, always look for relationships between angles (like exterior angles or angles opposite equal sides) to establish side relationships.

 

Question 17. Diagonals PR and QS of a quadrilateral PQRS intersect each other at O. Prove that:
(i) PQ + QR + RS + SP > PR + QS
(ii) PQ + QR + RS + SP < 2 (PR + QS)
Answer: We know that the sum of any two sides of a triangle is always greater than its third side. This is called the Triangle Inequality Theorem. (i) For proving PQ + QR + RS + SP > PR + QS: Apply the triangle inequality to the four triangles formed by the diagonals: In \( \triangle PQR \): \( PQ + QR > PR \) ... (1) In \( \triangle RSP \): \( RS + SP > PR \) ... (2) Adding (1) and (2): \( PQ + QR + RS + SP > PR + PR \)
\( \implies PQ + QR + RS + SP > 2PR \) ... (3) Similarly, apply the triangle inequality for the other diagonal QS: In \( \triangle PQS \): \( PQ + SP > QS \) ... (4) In \( \triangle RQS \): \( QR + RS > QS \) ... (5) Adding (4) and (5): \( PQ + SP + QR + RS > QS + QS \)
\( \implies PQ + QR + RS + SP > 2QS \) ... (6) Now, add inequality (3) and (6): \( (PQ + QR + RS + SP) + (PQ + QR + RS + SP) > 2PR + 2QS \)
\( \implies 2 (PQ + QR + RS + SP) > 2 (PR + QS) \) Divide both sides by 2:
\( \implies PQ + QR + RS + SP > PR + QS \) Hence proved for part (i). (ii) For proving PQ + QR + RS + SP < 2 (PR + QS): Consider the four smaller triangles formed by the intersection point O: In \( \triangle OPQ \): \( OP + OQ > PQ \) ... (a) In \( \triangle QOR \): \( OQ + OR > QR \) ... (b) In \( \triangle ROS \): \( OR + OS > RS \) ... (c) In \( \triangle SOP \): \( OS + OP > SP \) ... (d) Add all these four inequalities (a), (b), (c), and (d): \( (OP + OQ) + (OQ + OR) + (OR + OS) + (OS + OP) > PQ + QR + RS + SP \)
\( \implies OP + OQ + OQ + OR + OR + OS + OS + OP > PQ + QR + RS + SP \)
\( \implies 2OP + 2OQ + 2OR + 2OS > PQ + QR + RS + SP \)
\( \implies 2(OP + OQ + OR + OS) > PQ + QR + RS + SP \) Notice that \( (OP + OR) \) forms the diagonal PR, and \( (OQ + OS) \) forms the diagonal QS. So, we can write:
\( \implies 2((OP + OR) + (OQ + OS)) > PQ + QR + RS + SP \)
\( \implies 2(PR + QS) > PQ + QR + RS + SP \) This can be rewritten as:
\( \implies PQ + QR + RS + SP < 2(PR + QS) \) Hence proved for part (ii). The sum of all sides is both greater than the sum of diagonals and less than twice the sum of diagonals.
In simple words: This problem uses the simple rule that any two sides of a triangle are longer than the third side. We apply this rule many times to different triangles inside the quadrilateral to show how the total length of all sides compares to the lengths of the diagonals.

🎯 Exam Tip: When dealing with quadrilaterals and diagonals, always remember to break it down into smaller triangles and apply the triangle inequality theorem (a+b>c) repeatedly. Identify the relevant triangles carefully.

 

Question 18. Prove that any two sides of a triangle are together greater than the third side.
Answer: Given: A triangle ABC. To prove: (i) AB + AC > BC (ii) AC + BC > AB (iii) BC + AB > AC Construction: Extend side BA to a point D such that AD = AC. Then, join C to D. Proof: In \( \triangle ADC \), since we constructed AD = AC, it means that \( \triangle ADC \) is an isosceles triangle. In an isosceles triangle, the angles opposite to the equal sides are also equal. So, \( \angle ACD = \angle ADC \). From the figure, we can see that \( \angle BCD = \angle BCA + \angle ACD \). This clearly means that \( \angle BCD > \angle ACD \). Since \( \angle ACD = \angle ADC \), we can say that \( \angle BCD > \angle ADC \). Now, consider the larger triangle \( \triangle BCD \). In \( \triangle BCD \), we have established that \( \angle BCD > \angle BDC \) (which is the same as \( \angle ADC \)). In any triangle, the side opposite the greater angle is longer. So, the side opposite \( \angle BCD \) (which is BD) must be greater than the side opposite \( \angle BDC \) (which is BC).
\( \implies BD > BC \) From our construction, we know that BD is made up of BA + AD. So, \( BA + AD > BC \). Since we constructed AD = AC, we can substitute AC for AD:
\( \implies BA + AC > BC \) This proves the first part (AB + AC > BC). Similarly, by extending other sides and making similar constructions, we can prove the other two inequalities: AC + BC > AB and BC + AB > AC. The triangle inequality is a fundamental property that ensures a triangle can be formed.
In simple words: To make a triangle, if you pick any two sides, their combined length must be more than the length of the third side. If it's not, the two shorter sides won't be able to meet. We can show this by drawing an extra line to make an isosceles triangle, which helps compare the angles and sides.

🎯 Exam Tip: When proving the triangle inequality, the key is the construction step where you extend one side and make an isosceles triangle. This creates equal angles that help relate the sum of two sides to the third side.

 

Question 19. Prove that in a triangle, the difference of any two sides is less than the third side.
Answer: Given: A triangle ABC with sides AB, BC, and AC. Assume AC > AB without loss of generality. To prove: \( AC - AB < BC \) Construction: On side AC, cut off a segment AD such that AD = AB. Then, join B to D. Proof: In \( \triangle ABD \), since AD = AB by construction, \( \triangle ABD \) is an isosceles triangle. Therefore, the angles opposite the equal sides are equal: \( \angle ABD = \angle ADB \). We know that an exterior angle of a triangle is greater than either of its interior opposite angles. Consider \( \triangle BCD \). The angle \( \angle ADB \) is an exterior angle to \( \triangle BCD \). So, \( \angle ADB > \angle CBD \) and \( \angle ADB > \angle BCD \). Specifically, we will use \( \angle ADB > \angle CBD \). Since \( \angle ADB = \angle ABD \), we can say that \( \angle ABD > \angle CBD \). This implies that in \( \triangle BCD \), the angle \( \angle BDC \) (which is \( \angle ADB \)) is greater than \( \angle DBC \). In \( \triangle BCD \), the side opposite \( \angle BDC \) is BC, and the side opposite \( \angle DBC \) is CD. Since \( \angle BDC > \angle DBC \), it follows that \( BC > CD \). From our construction, CD is the difference between AC and AD. \( CD = AC - AD \) Since we set AD = AB, we can substitute AB for AD: \( CD = AC - AB \) Now, substitute this back into our inequality \( BC > CD \):
\( \implies BC > AC - AB \) This proves that the difference between any two sides of a triangle is less than the third side. This is an important part of understanding triangle properties.
In simple words: This rule means that if you take the length of one side and subtract the length of another side, the answer will always be smaller than the length of the third side. For example, a 10cm side and a 3cm side can't have a third side that is shorter than 7cm. We prove this by drawing a line inside the triangle to create a smaller isosceles triangle, helping us compare the lengths.

🎯 Exam Tip: This proof often uses a construction similar to the triangle inequality proof, but instead of extending a side, you cut a segment on a longer side. This allows you to use properties of isosceles triangles and exterior angles to show the inequality.