RBSE Solutions Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Exercise 7.1

Question 1. In ∆ABC and APQR, ∠A = ∠Q and ∠B = ∠R. Then which side of APQR is equal to side AB of ∆ABC. So that both triangles become congruent. Give reason for your answer.
Answer: For triangles \( \triangle ABC \) and \( \triangle PQR \) to be congruent, given that \( \angle A = \angle Q \) and \( \angle B = \angle R \), the side AB of \( \triangle ABC \) should be equal to QR. This means \( AB = QR \). When this condition is met, the triangles \( \triangle ABC \) and \( \triangle PQR \) become congruent by the Angle-Side-Angle (ASA) rule. ASA congruence requires the side to be included between the two equal angles. Since \( \angle A \) matches \( \angle Q \) and \( \angle B \) matches \( \angle R \), the side included between \( \angle A \) and \( \angle B \) is AB, and the side included between \( \angle Q \) and \( \angle R \) is QR.
In simple words: To make the two triangles exactly the same size and shape, the side AB from the first triangle must be the same length as side QR from the second triangle. This is because we need the side to be "sandwiched" between the two angles that are already known to be equal.

🎯 Exam Tip: When using the ASA (Angle-Side-Angle) congruence criterion, always ensure that the equal side is positioned exactly between the two equal angles in both triangles. If the side is not included, a different congruence rule might apply, or the triangles might not be congruent.

Question 2. In triangles, ABC and PQR, ∠A = ∠Q and ∠B = ∠R then which side of APQR is equal to side BC of ∆ABC, so that both triangles congruent? Give reasons for your answer.
Answer: In \( \triangle ABC \) and \( \triangle PQR \), we are given \( \angle A = \angle Q \) and \( \angle B = \angle R \).
\( \implies \) Since two angles are equal, the third angle must also be equal: \( \angle C = \angle P \). This is because the sum of angles in any triangle is always 180 degrees. For side BC of \( \triangle ABC \) to be equal to a side in \( \triangle PQR \) for congruence, BC is included between \( \angle B \) and \( \angle C \). Therefore, the corresponding side in \( \triangle PQR \) must be PR (included between \( \angle R \) and \( \angle P \)). Thus, \( BC = PR \) would make the triangles congruent by ASA.
In simple words: If two angles of two triangles are equal, then their third angles are also equal. For side BC to match a side in the other triangle, it should be equal to side PR. This makes the triangles congruent by the ASA rule.

🎯 Exam Tip: Remember that if two angles of one triangle are equal to two angles of another triangle, their third angles must also be equal. This property is often useful in congruence proofs, allowing you to use ASA even if the initially given angles are not immediately adjacent to the side.

Question 3. If two sides and one angle of a triangle are equal to two sides and the angle of other then both the triangles must congruent to each other. Is this statement true?
Answer: No, this statement is not true. For any two triangles to be congruent, it is essential that if two sides and one angle are equal, the angle must be the *included* angle (the angle between the two sides). This is known as the SAS (Side-Angle-Side) congruence criterion. If the angle is not the included angle, the triangles might not be congruent. For example, two triangles could have two equal sides and one equal non-included angle but still have different shapes and sizes.
In simple words: No, this is false. For triangles to be congruent when two sides and one angle are equal, that angle must be exactly between those two sides.

🎯 Exam Tip: Clearly distinguish between SAS (Side-Angle-Side) and ASS/SSA. SAS is a valid congruence rule, but ASS/SSA (where the angle is not included between the two sides) is not, as it can lead to ambiguous cases where two different triangles can be formed.

Question 4. If two angles and one side of a triangle are equal to two angles and one side of other triangle then both A's must congruent to each other. Is this statement true?
Answer: No, this statement is not true. For two triangles to be congruent when two angles and one side are equal, the side must be *included* between the two angles. This is the ASA (Angle-Side-Angle) congruence criterion. If the side is not the included side, the triangles might not be congruent. For example, the AAS (Angle-Angle-Side) criterion also works, but the statement as given isn't precise enough to guarantee congruence without specifying the side's position.
In simple words: No, this is false. For triangles to be congruent, the equal side must be located between the two equal angles.

🎯 Exam Tip: Understand the difference between ASA (Angle-Side-Angle) and AAS (Angle-Angle-Side). Both are valid congruence criteria, but the problem's phrasing implies a specific setup, which needs to be carefully checked.

Question 5. If ∆ABC = ARPQ is given. It is true to say BC = QR? and why?
Answer: No, it is not true to say \( BC = QR \).
\( \because \triangle ABC \cong \triangle RPQ \) (given congruence, assuming 'equal' means congruent). When two triangles are congruent, their corresponding parts are equal (CPCTC - Corresponding Parts of Congruent Triangles are Congruent). Based on the order of the vertices in the congruence statement, if \( \triangle ABC \cong \triangle RPQ \), then:
\( \implies \) A corresponds to R
\( \implies \) B corresponds to P
\( \implies \) C corresponds to Q So, side BC corresponds to side PQ. Therefore, \( BC \) should be equal to \( PQ \), not \( QR \). Congruence statements like \( \triangle ABC \cong \triangle RPQ \) imply a specific mapping of vertices.
In simple words: No, BC is not equal to QR. When triangles are congruent, their matching parts are equal. In this case, BC matches with PQ because of how the letters are lined up in the congruence statement.

🎯 Exam Tip: Always pay close attention to the order of vertices in a congruence statement (e.g., \( \triangle ABC \cong \triangle PQR \)). This order tells you exactly which angles and sides correspond to each other, which is crucial for applying CPCTC correctly.

Question 6. If APQR = AEDF then is it true to say PR = EF? Give reasons for your answer.
Answer: Yes, it is true to say that \( PR = EF \).
\( \because \triangle PQR \cong \triangle EDF \) (given congruence, assuming 'equal' means congruent). When two triangles are congruent, their corresponding parts are equal. Based on the order of the vertices in the congruence statement:
\( \implies \) P corresponds to E
\( \implies \) Q corresponds to D
\( \implies \) R corresponds to F So, side PR corresponds to side EF. Therefore, \( PR \) is equal to \( EF \) because they are corresponding sides of congruent triangles. This is a direct application of the CPCTC rule.
In simple words: Yes, PR is equal to EF. When triangles are congruent, their matching sides are also equal. Here, PR matches EF because of the way the triangles are named as being congruent.

🎯 Exam Tip: The principle of Corresponding Parts of Congruent Triangles are Congruent (CPCTC) is fundamental. Once congruence is established, any pair of corresponding sides or angles can be declared equal based on the vertex order in the congruence statement.

Question 7. In figure, the diagonal AC of a quadrilateral ABCD bisects the angle A and C. Prove that AB = AD and CB = CD.
Answer: Consider the triangles \( \triangle ABC \) and \( \triangle ADC \). Given that the diagonal AC bisects angle A, we have \( \angle BAC = \angle DAC \). Given that the diagonal AC bisects angle C, we have \( \angle BCA = \angle DCA \). The side AC is common to both triangles. So, in \( \triangle ABC \) and \( \triangle ADC \): \( \angle BAC = \angle DAC \) (Angle, given) \( AC = AC \) (Side, common) \( \angle BCA = \angle DCA \) (Angle, given)
\( \implies \triangle ABC \cong \triangle ADC \) (by ASA congruency property). Since the triangles are congruent, their corresponding parts are equal (CPCTC).
\( \implies AB = AD \) and \( CB = CD \). Hence, it is proved.
In simple words: We can prove that triangles ABC and ADC are congruent. Since diagonal AC cuts angles A and C in half, and AC is a common side, the triangles are congruent by the ASA rule. Because they are congruent, their matching sides AB equals AD, and CB equals CD.

🎯 Exam Tip: When proving parts of a figure are equal, look for congruent triangles first. Identify common sides, given bisected angles, or parallel lines to establish congruence using criteria like SSS, SAS, ASA, AAS, or RHS.

Question 8. In figure, ADBC is a quadrilateral in which ∠ABC = ∠ABD and BC = BD then show that ∆ABC = ∆ABD.
Answer: Consider the triangles \( \triangle ABC \) and \( \triangle ABD \). We are given the following information: \( \angle ABC = \angle ABD \) (Angle, given) \( BC = BD \) (Side, given) \( AB = AB \) (Side, common to both triangles)
\( \implies \triangle ABC \cong \triangle ABD \) (By SAS congruency property). Hence, it is proved that the two triangles are congruent. The SAS rule is used because the equal angle is between the two equal sides.
In simple words: In the two triangles ABC and ABD, we are told that one angle and two sides are equal in a specific order (Side-Angle-Side). This means the triangles are congruent by the SAS rule.

🎯 Exam Tip: For SAS congruence, it's critical that the equal angle is the *included* angle, meaning it is formed by the two equal sides. If the angle is not between the sides, SAS cannot be used.

Question 9. In figure, AB || CD and AD || BC then show that ∆ADB = ∆BCD.
Answer: Consider the triangles \( \triangle ADB \) and \( \triangle BCD \). Given that \( AB \parallel CD \):
\( \implies \angle ABD = \angle CDB \) (Alternate interior angles). Let's call these \( \angle 1 \) and \( \angle 2 \) respectively. Also, given that \( AD \parallel BC \):
\( \implies \angle ADB = \angle CBD \) (Alternate interior angles). Let's call these \( \angle 3 \) and \( \angle 4 \) respectively. The side BD is common to both triangles: \( BD = BD \) (Common side). So, in \( \triangle ADB \) and \( \triangle BCD \): \( \angle ABD = \angle CDB \) (Angle) \( BD = BD \) (Side) \( \angle ADB = \angle CBD \) (Angle)
\( \implies \triangle ADB \cong \triangle BCD \) (by ASA congruency property). Hence, it is proved that the two triangles are congruent.
In simple words: Since AB is parallel to CD, and AD is parallel to BC, we can find pairs of equal alternate angles. Also, BD is a common side. Using the Angle-Side-Angle (ASA) rule, we can show that triangle ADB is congruent to triangle BCD.

🎯 Exam Tip: When dealing with parallel lines, always look for alternate interior angles or corresponding angles, as these provide equal angle pairs crucial for proving triangle congruence using ASA or AAS criteria.

Question 10. In figure, if AB || CD and E is the mid-point of AC, then show that E is the mid-point of BD.
Answer: Consider the triangles \( \triangle AEB \) and \( \triangle DEC \). Given that \( AB \parallel CD \):
\( \implies \angle EAB = \angle ECD \) (Alternate interior angles). In the figure, this corresponds to \( \angle 1 = \angle 2 \). Also, \( \angle AEB = \angle DEC \) (Vertically opposite angles). In the figure, this corresponds to \( \angle 3 = \angle 4 \). Given that E is the mid-point of AC:
\( \implies AE = EC \) (Side, by definition of midpoint). So, in \( \triangle AEB \) and \( \triangle DEC \): \( \angle EAB = \angle ECD \) (Angle) \( AE = EC \) (Side) \( \angle AEB = \angle DEC \) (Angle)
\( \implies \triangle AEB \cong \triangle DEC \) (by ASA congruency property). Since the triangles are congruent, their corresponding parts are equal (CPCTC).
\( \implies DE = EB \).
\( \implies \) This means that E is the mid-point of BD. Hence, it is proved.
In simple words: We compare triangles AEB and DEC. Since AB is parallel to CD, we find equal alternate angles. Also, E is the midpoint of AC, and vertically opposite angles are equal. So, the triangles are congruent by the ASA rule. Because they are congruent, their matching sides DE and EB must be equal, which proves E is the midpoint of BD.

🎯 Exam Tip: Proving two segments are equal often involves showing that they are corresponding parts of congruent triangles. Carefully identify the relevant triangles and use given information (like parallel lines or midpoints) to establish congruence.