RBSE Solutions Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Exercise 7.2

Question 1. In figure, AB = AC and \( \angle B = 58^\circ \), then find the value of \( \angle A \).Answer: In triangle ABC, since AB = AC, the angles opposite to these sides must be equal. This means \( \angle B = \angle C \). Given that \( \angle B = 58^\circ \), then \( \angle C \) is also \( 58^\circ \). The sum of angles in a triangle is \( 180^\circ \). So, to find \( \angle A \), we subtract the sum of \( \angle B \) and \( \angle C \) from \( 180^\circ \).
\( \angle A = 180^\circ - (\angle B + \angle C) \)
\( \angle A = 180^\circ - (58^\circ + 58^\circ) \)
\( \angle A = 180^\circ - 116^\circ \)
\( \angle A = 64^\circ \)
In simple words: If two sides of a triangle are the same length, the angles opposite those sides are also equal. Since two angles are \( 58^\circ \) each, the third angle is found by taking them away from \( 180^\circ \).

🎯 Exam Tip: Remember the property that "angles opposite to equal sides are equal" in an isosceles triangle. Always show the sum of angles in a triangle equals 180 degrees.

Question 2. In figure, AD = BD and \( \angle C = \angle E \). Prove that BC = AE.Answer: We need to prove that BC = AE. We are given AD = BD and \( \angle C = \angle E \). Let's consider triangles \( \Delta ADE \) and \( \Delta BDC \).
1. \( AD = BD \) (Given)
2. \( \angle ADE = \angle BDC \) (These are vertically opposite angles, so they are equal)
3. \( \angle E = \angle C \) (Given)
Now, we have two angles and a non-included side equal in both triangles. According to the AAS (Angle-Angle-Side) congruence rule, if two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the triangles are congruent.
So, \( \Delta ADE \cong \Delta BDC \) (By AAS congruence rule)
When two triangles are congruent, their corresponding parts are equal (CPCTC).
Therefore, \( AE = BC \) (By CPCTC)
Hence, it is proved that BC = AE.
In simple words: We compare two small triangles. Since they have matching angles and one matching side, they are exactly the same shape and size. Because they are the same, their other matching sides (BC and AE) must also be equal.

🎯 Exam Tip: When proving congruence, clearly state the pairs of equal sides and angles. Remember to identify the correct congruence rule (like AAS, SAS, SSS, ASA, RHS) and use CPCTC to conclude equality of corresponding parts.

Question 3. If the bisector of an angle of a triangle also bisects the opposite side, show that the triangle is isosceles.
Answer: Let's consider a triangle ABC. Let AD be the bisector of \( \angle A \), so \( \angle BAD = \angle CAD \). We are also given that AD bisects the opposite side BC, which means D is the midpoint of BC, so \( BD = DC \).
To prove that the triangle ABC is isosceles, we need to show that \( AB = AC \).
1. Extend AD to a point E such that AD = DE.
2. Join point E to C.
Now, let's look at \( \Delta ABD \) and \( \Delta ECD \):
* \( BD = DC \) (Given, as AD bisects BC)
* \( AD = DE \) (By construction)
* \( \angle ADB = \angle EDC \) (These are vertically opposite angles, so they are equal)
Therefore, \( \Delta ABD \cong \Delta ECD \) (By SAS congruence rule)
From this congruence, we can say that:
* \( AB = EC \) (By CPCTC)
* \( \angle BAD = \angle CED \) (By CPCTC)
We know that AD is the bisector of \( \angle A \), so \( \angle BAD = \angle CAD \).
Since \( \angle BAD = \angle CED \) and \( \angle BAD = \angle CAD \), it means \( \angle CAD = \angle CED \).
Now, consider \( \Delta ACE \). In this triangle, since \( \angle CAD = \angle CED \), the sides opposite to these equal angles must also be equal.
Therefore, \( AC = EC \).
We have already shown that \( AB = EC \) and now we found \( AC = EC \).
This implies \( AB = AC \).
Since two sides of \( \Delta ABC \) are equal, \( \Delta ABC \) is an isosceles triangle. This construction and proof show how the conditions lead to the triangle being isosceles.
In simple words: If a line cuts an angle into two equal parts and also cuts the opposite side into two equal parts, then the two sides of the triangle next to that angle must be equal. This makes the triangle an "isosceles" triangle.

🎯 Exam Tip: This type of proof often requires an auxiliary construction (like extending a line). Clearly state your construction and then use congruence rules to find equal sides or angles. Remember that angles opposite equal sides are equal, and sides opposite equal angles are equal.

Question 4. If the bisector of an angle of a triangle also bisects the opposite side, show that the triangle is isosceles.Answer: Let \( \Delta ABC \) be a triangle. Let AD be the bisector of \( \angle BAC \), which means \( \angle BAD = \angle CAD \).
The problem statement is slightly different here, often implying that the bisector is also perpendicular to the opposite side. If we assume, as the solution seems to, that AD is perpendicular to BC (i.e., \( AD \perp BC \)), then \( \angle ADB = \angle ADC = 90^\circ \).
Now, let's consider \( \Delta ABD \) and \( \Delta ACD \):
1. \( \angle BAD = \angle CAD \) (Given, as AD bisects \( \angle BAC \))
2. \( AD = AD \) (Common side to both triangles)
3. \( \angle ADB = \angle ADC = 90^\circ \) (Given assumption: AD is perpendicular to BC)
Therefore, \( \Delta ABD \cong \Delta ACD \) (By ASA congruence rule: Angle-Side-Angle)
Since the triangles are congruent, their corresponding parts are equal (CPCTC).
This means \( AB = AC \) (By CPCTC)
Since two sides of \( \Delta ABC \) are equal, \( \Delta ABC \) is an isosceles triangle. This is a common way to prove a triangle is isosceles under these conditions.
In simple words: When a line divides an angle in a triangle into two equal halves, and also meets the opposite side at a right angle, then the two sides of the triangle that form the divided angle are equal. This makes the triangle special, calling it "isosceles."

🎯 Exam Tip: Some geometry problems can have slightly different interpretations depending on implicit conditions. When a bisector also bisects the opposite side, it usually implies it's perpendicular. Clearly state any assumptions made based on the provided solution or diagram.

Question 5. In figure, AB = AC and BE = CD. Prove that AD = AE.Answer: We are given that AB = AC and BE = CD. We need to prove AD = AE.
First, let's look at the given equality BE = CD.
If we subtract the segment DE from both BE and CD, we get:
\( BE - DE = CD - DE \)
This simplifies to \( BD = CE \) (Let's call this Equation 1)
Now, let's consider \( \Delta ABD \) and \( \Delta ACE \):
1. \( AB = AC \) (Given)
2. \( \angle ABD = \angle ACE \) (Since AB = AC in \( \Delta ABC \), the angles opposite to these equal sides are also equal, so \( \angle B = \angle C \))
3. \( BD = CE \) (Proved in Equation 1)
Therefore, by the SAS (Side-Angle-Side) congruence rule, if two sides and the included angle of one triangle are equal to the corresponding sides and included angle of another triangle, then the triangles are congruent.
So, \( \Delta ABD \cong \Delta ACE \) (By SAS congruence rule)
Since the triangles are congruent, their corresponding parts are equal (CPCTC).
Therefore, \( AD = AE \) (By CPCTC)
Hence, it is proved.
In simple words: We start with what we know about the lengths of the lines. By taking a common part out, we find that some smaller parts are also equal. Then, by comparing two triangles and using the Side-Angle-Side rule, we show they are identical. This makes their remaining matching sides equal.

🎯 Exam Tip: When parts of lines are given as equal, try to manipulate those equations by adding or subtracting common segments to find new equal parts that can be used in triangle congruence proofs. Remember that in an isosceles triangle, angles opposite equal sides are equal.

Question 6. Two points E and F of the sides AD and BC respectively of a square ABCD such that AF = BE then prove that.
(i) \( \angle BAF = \angle ABE \)
(ii) BF = AEAnswer: We are given a square ABCD, with points E on AD and F on BC such that AF = BE. We need to prove (i) \( \angle BAF = \angle ABE \) and (ii) BF = AE.
Let's consider \( \Delta ABE \) and \( \Delta BAF \).
1. \( AB = BA \) (Common side to both triangles)
2. \( \angle BAE = 90^\circ \) (Angle of a square ABCD, \( \angle A \) is \( 90^\circ \))
3. \( \angle ABF = 90^\circ \) (Angle of a square ABCD, \( \angle B \) is \( 90^\circ \))
4. \( AF = BE \) (Given, these are the hypotenuses of the right-angled triangles)
Since both triangles \( \Delta ABE \) and \( \Delta BAF \) are right-angled triangles, we can use the RHS (Right angle-Hypotenuse-Side) congruence rule.
Therefore, \( \Delta ABE \cong \Delta BAF \) (By RHS congruence rule)
Now, because the triangles are congruent, their corresponding parts are equal (CPCTC).
(i) From CPCTC, \( \angle BAF = \angle ABE \) (Corresponding angles)
(ii) From CPCTC, \( BF = AE \) (Corresponding sides)
Hence, both statements are proved.
In simple words: Since we have a square, we know certain angles are \( 90^\circ \). By comparing two triangles using the Right angle, Hypotenuse, and Side rule, we show they are identical. This means their matching angles and sides are also equal, proving both parts of the question.

🎯 Exam Tip: When dealing with squares or rectangles, remember that all angles are 90 degrees and opposite sides are equal. The RHS congruence rule is very useful for proving congruence in right-angled triangles if the hypotenuse and one side are equal.

Question 7. AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.Answer: We are given that AD and BC are equal perpendiculars to the line segment AB. This means \( AD \perp AB \) and \( BC \perp AB \), and \( AD = BC \). We need to show that CD bisects AB, which means O is the midpoint of AB, so \( OA = OB \).
Let O be the point where CD and AB intersect.
Consider \( \Delta OAD \) and \( \Delta OBC \):
1. \( \angle DAO = \angle CBO = 90^\circ \) (Since AD and BC are perpendicular to AB)
2. \( \angle AOD = \angle BOC \) (These are vertically opposite angles, so they are equal)
3. \( AD = BC \) (Given that they are equal perpendiculars)
Therefore, by the AAS (Angle-Angle-Side) congruence rule, if two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the triangles are congruent.
So, \( \Delta OAD \cong \Delta OBC \) (By AAS congruence rule)
Since the triangles are congruent, their corresponding parts are equal (CPCTC).
Therefore, \( OA = OB \) (By CPCTC)
Since \( OA = OB \), it means that O is the midpoint of AB. This shows that the line segment CD bisects AB. This proof uses the concept of congruent triangles effectively.
In simple words: We have two lines standing straight up from a base line, and they are the same height. When we draw a line connecting their tops, it cuts the base line exactly in the middle. We prove this by showing that the two small triangles formed are identical.

🎯 Exam Tip: When proving that a line bisects another, aim to show that the point of intersection creates two equal segments. Look for pairs of triangles that share this point and apply congruence rules like AAS or ASA.

Question 8. In an isosceles triangle ABC with AB = AC, the bisectors of \( \angle B \) and \( \angle C \) meet at O. Produce BO to M. Show that \( \angle MOC = \angle ABC \).
Answer: We are given an isosceles triangle ABC where \( AB = AC \). Therefore, the angles opposite to these equal sides are also equal, so \( \angle ABC = \angle ACB \).
BO is the bisector of \( \angle ABC \), so \( \angle OBC = \frac{1}{2} \angle ABC \).
CO is the bisector of \( \angle ACB \), so \( \angle OCB = \frac{1}{2} \angle ACB \).
Since \( \angle ABC = \angle ACB \), it means \( \angle OBC = \angle OCB \).
Now, consider \( \Delta OBC \). We know that the exterior angle of a triangle is equal to the sum of its two interior opposite angles.
Here, \( \angle MOC \) is an exterior angle to \( \Delta OBC \).
So, \( \angle MOC = \angle OBC + \angle OCB \).
Since \( \angle OBC = \frac{1}{2} \angle ABC \) and \( \angle OCB = \frac{1}{2} \angle ACB \), and \( \angle ABC = \angle ACB \), we can write:
\( \angle MOC = \frac{1}{2} \angle ABC + \frac{1}{2} \angle ABC \)
\( \angle MOC = \left( \frac{1}{2} + \frac{1}{2} \right) \angle ABC \)
\( \angle MOC = 1 \times \angle ABC \)
\( \angle MOC = \angle ABC \)
Hence, it is proved that \( \angle MOC = \angle ABC \). This property is a direct application of angle bisectors and exterior angle theorem in isosceles triangles.
In simple words: In a special triangle where two sides are equal, the lines that cut the bottom two angles in half meet at a point. If we extend one of these lines, the angle formed outside is equal to one of the original big angles of the triangle.

🎯 Exam Tip: Always remember the properties of isosceles triangles (equal sides imply equal opposite angles). The exterior angle theorem (exterior angle equals sum of two opposite interior angles) is a powerful tool in geometry proofs. Clearly define angle bisectors in your steps.

Question 9. Line l is the bisector of an \( \angle A \). B is any point on l. BP and BQ are perpendicular from B to the arms of \( \angle A \) (see figure). Show that
(i) \( \Delta APB = \Delta AQB \)
(ii) BP = BQ or B is equidistant from the arms of \( \angle A \).Answer: We are given that line 'l' bisects \( \angle A \), so any point on 'l' (like B) is such that \( \angle BAP = \angle BAQ \). BP and BQ are perpendiculars from B to the arms of \( \angle A \), meaning \( \angle BPA = 90^\circ \) and \( \angle BQA = 90^\circ \).
(i) Let's consider \( \Delta APB \) and \( \Delta AQB \):
1. \( \angle BPA = \angle BQA = 90^\circ \) (Given that BP and BQ are perpendiculars)
2. \( \angle BAP = \angle BAQ \) (Given, as line 'l' is the angle bisector of \( \angle A \))
3. \( AB = AB \) (Common side to both triangles)
Therefore, by the AAS (Angle-Angle-Side) congruence rule, if two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the triangles are congruent.
So, \( \Delta APB \cong \Delta AQB \) (By AAS congruence rule)
(ii) Since \( \Delta APB \cong \Delta AQB \) (Proved in part i), their corresponding parts are equal (CPCTC).
Therefore, \( BP = BQ \) (By CPCTC)
This means that point B is equidistant from the arms of \( \angle A \). This is a fundamental property of angle bisectors: any point on an angle bisector is equidistant from the arms of the angle.
In simple words: We have a line cutting an angle in half, and a point on that line. If we draw straight lines from this point to each side of the angle, making a \( 90^\circ \) corner, the two small triangles formed are identical. This means the two straight lines drawn are the same length, showing the point is equally far from both sides of the angle.

🎯 Exam Tip: The AAS congruence rule is powerful when you have two angles and a non-included side. Remember that the definition of an angle bisector is that any point on it is equidistant from the angle's arms – this is a key geometric property to state clearly.

Question 10. In figure, AC = AE, AB = AD and \( \angle BAD = \angle EAC \), show that BC = DE.Answer: We are given \( AC = AE \), \( AB = AD \), and \( \angle BAD = \angle EAC \). We need to show that BC = DE.
We are given \( \angle BAD = \angle EAC \).
Let's add \( \angle DAC \) to both sides of this equality:
\( \angle BAD + \angle DAC = \angle EAC + \angle DAC \)
This sum simplifies to:
\( \angle BAC = \angle DAE \) (Let's call this Equation 1)
Now, let's consider \( \Delta ABC \) and \( \Delta ADE \):
1. \( AB = AD \) (Given)
2. \( \angle BAC = \angle DAE \) (Proved in Equation 1)
3. \( AC = AE \) (Given)
Therefore, by the SAS (Side-Angle-Side) congruence rule, if two sides and the included angle of one triangle are equal to the corresponding sides and included angle of another triangle, then the triangles are congruent.
So, \( \Delta ABC \cong \Delta ADE \) (By SAS congruence rule)
Since the triangles are congruent, their corresponding parts are equal (CPCTC).
Therefore, \( BC = DE \) (By CPCTC)
Hence, it is proved.
In simple words: We know some side lengths are equal and a small part of an angle is equal. By adding a common angle, we make the whole angles equal. Then, by comparing two triangles, we find they are identical in shape and size. This makes the remaining matching sides also equal.

🎯 Exam Tip: In problems involving overlapping triangles, adding or subtracting common angles/segments is a key technique to create equal parts needed for congruence. Always specify the congruence criterion used (SAS, ASA, SSS, RHS, AAS).

Question 11. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that:
(i) \( \Delta AMC \cong \Delta BMD \)
(ii) \( \angle DBC \) is a right angle
(iii) \( \Delta DBC \cong \Delta ACB \)
(iv) \( CM = \frac {1}{2} AB \)Answer: We are given a right-angled triangle ABC, with \( \angle C = 90^\circ \). M is the midpoint of the hypotenuse AB, and CM is produced to D such that DM = CM.
(i) To show \( \Delta AMC \cong \Delta BMD \):
Let's consider \( \Delta AMC \) and \( \Delta BMD \):
1. \( AM = BM \) (Given that M is the midpoint of AB)
2. \( CM = DM \) (Given by construction)
3. \( \angle AMC = \angle BMD \) (These are vertically opposite angles, so they are equal)
Therefore, by the SAS (Side-Angle-Side) congruence rule,
\( \Delta AMC \cong \Delta BMD \)
(ii) To show \( \angle DBC \) is a right angle:
Since \( \Delta AMC \cong \Delta BMD \) (Proved in part i), their corresponding parts are equal (CPCTC).
Therefore, \( AC = DB \) and \( \angle CAM = \angle DBM \).
\( \angle CAM \) and \( \angle DBM \) are alternate interior angles for lines AC and DB with transversal AB. Since these alternate interior angles are equal, it implies that \( AC \parallel DB \).
Now, consider the parallel lines AC and DB, with BC as a transversal.
The sum of consecutive interior angles is \( 180^\circ \).
So, \( \angle ACB + \angle DBC = 180^\circ \)
We are given that \( \angle ACB = 90^\circ \) (Since \( \Delta ABC \) is right-angled at C).
Therefore, \( 90^\circ + \angle DBC = 180^\circ \)
\( \angle DBC = 180^\circ - 90^\circ \)
\( \angle DBC = 90^\circ \)
Hence, \( \angle DBC \) is a right angle.
(iii) To show \( \Delta DBC \cong \Delta ACB \):
Let's consider \( \Delta DBC \) and \( \Delta ACB \):
1. \( DB = AC \) (Proved from CPCTC in part i: \( \Delta AMC \cong \Delta BMD \))
2. \( BC = CB \) (Common side to both triangles)
3. \( \angle DBC = \angle ACB = 90^\circ \) (Proved in part ii, and given for \( \angle ACB \))
Therefore, by the SAS (Side-Angle-Side) congruence rule,
\( \Delta DBC \cong \Delta ACB \)
(iv) To show \( CM = \frac {1}{2} AB \):
Since \( \Delta DBC \cong \Delta ACB \) (Proved in part iii), their corresponding parts are equal (CPCTC).
Therefore, \( CD = AB \).
We know that CD is formed by CM + MD.
And we are given that \( CM = DM \).
So, \( CD = CM + CM = 2CM \).
Since \( CD = AB \) and \( CD = 2CM \), we can write:
\( 2CM = AB \)
\( CM = \frac {1}{2} AB \)
Hence, it is proved that CM is half the length of AB. This shows that in a right-angled triangle, the median to the hypotenuse is half the length of the hypotenuse.
In simple words: This problem asks us to prove several things about a right triangle. First, two small triangles formed are identical. This helps us show that a newly formed angle is also a right angle. Then, we prove two larger triangles are identical, which finally helps us show that the line from the right angle to the middle of the longest side is exactly half the length of that longest side.

🎯 Exam Tip: This is a multi-part proof, so break it down step-by-step. Remember that each part builds on the previous one. Key concepts include midpoint definitions, vertically opposite angles, CPCTC, alternate interior angles for parallel lines, and the sum of consecutive interior angles.