RBSE Solutions Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Important Questions

Multiple Choice Questions

Question 1. In figure, if AB = AC and ∠B = 70°, the value of ∠A will be:
(a) 70°
(b) 40°
(c) 55°
(d) 90°
Answer: (b) 40°
In simple words: Since AB = AC, the angles opposite to these sides are equal, so ∠C is also 70°. Because all angles in a triangle add up to 180°, angle A will be 180° - 70° - 70°, which is 40°. This is a basic property of isosceles triangles.

🎯 Exam Tip: Remember that in an isosceles triangle, the angles opposite the equal sides are always equal. This is key for solving such problems.

Question 2. In figure, if AB = AC, then the value of ∠C is equal to:
(a) 60°
(b) 36°
(c) 72°
(d) 108°
Answer: (c) 72°
In simple words: Since AB = AC, the angles opposite these sides are equal. The angle at A is shown as 'x' and angle B is '2x'. So, angle C will also be '2x'. Because all angles in a triangle add up to 180°, we have x + 2x + 2x = 180°, which means 5x = 180°, so x = 36°. Therefore, angle C is 2x, which is 72°. The sum of angles in any triangle is always 180 degrees.

🎯 Exam Tip: When given algebraic expressions for angles in an isosceles triangle, set up an equation using the property that angles opposite equal sides are equal, and the sum of angles is 180°.

Question 4. In figure, AB = AC and AD ⊥ BC, then AD bisects the:
(a) ∠A
(b) Side BC
(c) ∠A and side BC
(d) None of these
Answer: (c) ∠A and side BC
In simple words: In an isosceles triangle where AB and AC are equal, the altitude (AD) drawn from the vertex A to the base BC not only forms a 90° angle with the base but also divides the vertex angle A into two equal halves and bisects the base BC. This is a special property of isosceles triangles.

🎯 Exam Tip: For an isosceles triangle, the altitude from the vertex angle to the base is also the angle bisector and the median to the base. This is a powerful property for proofs and calculations.

Question 5. ΔABC, shown in figure, in which AD = BD, and AC = DC and ∠C = 44°, then ∠A is:
(a) 68°
(b) 112°
(c) 34°
(d) 102°
Answer: (b) 112°
In simple words: In triangle ADC, since AC = DC, the angles opposite to them are equal, so ∠DAC = ∠D = 44°. In triangle ABD, since AD = BD, the angles opposite to them are also equal, so ∠ABD = ∠BAD. The angle ∠ADB is an exterior angle to triangle ADC, so ∠ADB = ∠DAC + ∠C = 44° + 44° = 88°. In triangle ABD, ∠BAD = ∠ABD = (180° - 88°)/2 = 46°. So, total angle A is ∠BAD + ∠DAC = 46° + 44° = 90°. There seems to be a discrepancy with the given solution (112°). Let's re-evaluate. If ∠D in ΔADC is 44°, then ∠ADC is 180 - 44 - 44 = 92°. Then ∠ADB = 180 - 92 = 88°. So ∠BAD in ΔABD is (180 - 88) / 2 = 46°. Total ∠A = ∠BAD + ∠CAD = 46 + 44 = 90°. The source answer 112° might be a typo or based on a different interpretation. Sticking to the diagram's explicit info: in ΔADC, since AC = DC, ∠CAD = ∠ADC. The solution provided in the source is (B) 112°. Let's try to achieve 112°. If ∠A = 112°, then in ΔADC, if AC = DC, and ∠C = 44°, then ∠CAD = ∠ADC. Let's assume ∠CAD = x. Then ∠ADC = x. So x + x + 44 = 180 => 2x = 136 => x = 68°. So ∠CAD = 68°. Now, for ΔABD, AD = BD. So ∠BAD = ∠B. If ∠CAD = 68°, then ∠A = ∠BAD + 68°. For the source answer to be 112°, ∠BAD would need to be 112 - 68 = 44°. If ∠BAD = 44°, then ∠B = 44°. So in ΔABD, ∠ADB = 180 - 44 - 44 = 92°. If ∠ADC = 68°, then this doesn't work. Let's follow the solution's logic: In ΔADC: AC = DC (given), ∠C = 44°. This means ∠CAD = ∠ADC (angles opposite equal sides). Let ∠CAD = x. Then ∠ADC = x. So, 2x + 44° = 180° (sum of angles in a triangle). 2x = 180° - 44° = 136°. x = 136° / 2 = 68°. So, ∠CAD = 68° and ∠ADC = 68°. Now consider ΔABD: AD = BD (given). This means ∠BAD = ∠ABD (angles opposite equal sides). ∠ADB is supplementary to ∠ADC. So ∠ADB = 180° - ∠ADC = 180° - 68° = 112°. In ΔABD: ∠BAD + ∠ABD + ∠ADB = 180°. Since ∠BAD = ∠ABD, let ∠BAD = y. 2y + 112° = 180°. 2y = 180° - 112° = 68°. y = 68° / 2 = 34°. So, ∠BAD = 34°. Finally, ∠A in ΔABC = ∠BAD + ∠CAD = 34° + 68° = 102°. This results in 102°, which is option (D). The source solution (B) 112° is incorrect. I will provide the correct calculation.
Answer: In ΔADC, we are given that AC = DC and ∠C = 44°. Since sides opposite to equal angles are equal, ∠CAD = ∠ADC. The sum of angles in a triangle is 180°, so ∠CAD + ∠ADC + ∠C = 180°. This means 2 * ∠CAD + 44° = 180°. So, 2 * ∠CAD = 136°, which gives ∠CAD = 68°. Also, ∠ADC = 68°. Next, consider ΔABD. We are given AD = BD. This means ∠BAD = ∠ABD (angles opposite equal sides). The angle ∠ADB is supplementary to ∠ADC, so ∠ADB = 180° - ∠ADC = 180° - 68° = 112°. In ΔABD, the sum of angles is 180°, so ∠BAD + ∠ABD + ∠ADB = 180°. Since ∠BAD = ∠ABD, we have 2 * ∠BAD + 112° = 180°. 2 * ∠BAD = 180° - 112° = 68°. ∠BAD = 68° / 2 = 34°. Therefore, the total angle ∠A in ΔABC = ∠CAD + ∠BAD = 68° + 34° = 102°. The correct option is (d). In simple words: First, use the fact that triangle ADC is isosceles (AC=DC) to find angles ∠CAD and ∠ADC. Then, use that ∠ADB and ∠ADC are on a straight line to find ∠ADB. Finally, use that triangle ABD is isosceles (AD=BD) to find ∠BAD, and add ∠CAD and ∠BAD to get the total angle ∠A. The sum of angles in a triangle is always 180 degrees.

🎯 Exam Tip: Always clearly identify isosceles triangles and use the property that angles opposite equal sides are equal. Remember that angles on a straight line add up to 180 degrees.

Question 7. In figure, AB = AC and ∠ABD = ∠ACD then ABDC is:
(a) Equilateral
(b) Isosceles
(c) Equiangular
(d) Scalene
Answer: (b) Isosceles
In simple words: Since AB = AC, triangle ABC is an isosceles triangle, so ∠ABC = ∠ACB. If ∠ABD = ∠ACD, it means that the parts of these base angles are also equal. This implies that the full angles at B and C, or the segments on the base are such that the overall shape maintains isosceles properties. Specifically, if two triangles are congruent (which can be derived from the given information), then their corresponding parts will be equal.

🎯 Exam Tip: Look for clues like equal sides or angles. When two sides of a triangle are equal, the triangle is isosceles, and the angles opposite those sides are equal.

Question 8. In figure, AB = AC and AD is the bisector of ∠BAC meet BC at D. If ∠BAC = 60° then ∠ADC is equal to:
(a) 30°
(b) 60°
(c) 90°.
Answer: (c) 90°
In simple words: In an isosceles triangle ABC with AB = AC, if AD bisects the angle at A (∠BAC), then AD also acts as the altitude to the base BC and as the median. This means AD is perpendicular to BC, so ∠ADC will be 90°. Since ∠BAC is 60°, this specific triangle would be equilateral if all angles were 60°, but we only know AB=AC. The property of the angle bisector in an isosceles triangle makes it perpendicular to the base.

🎯 Exam Tip: For an isosceles triangle, the angle bisector of the vertex angle (the angle between the two equal sides) is also the perpendicular bisector of the base.

Question 10. In the adjoining figure, if PQ = PR and QS = RT then ΔPST is equal to:
(a) Isosceles triangle
(b) Equilateral triangle
(c) Scalene triangle
(d) Isosceles right angled triangle
Answer: (a) Isosceles triangle
In simple words: If PQ = PR, then triangle PQR is an isosceles triangle, so ∠PQS = ∠PRQ. Since QS = RT, we can subtract equal parts from equal angles, or use congruence between triangles PQS and PRT (depending on if T is on QS extended or similar). If PQ = PR, then ΔPQR is isosceles. Also, given QS = RT. In ΔPST, if we prove PS=PT, then it will be isosceles. From the diagram, it is implied that T is on the extension of QR or similar, leading to PS = PT. The key is that equal sides lead to equal angles.

🎯 Exam Tip: When proving a triangle is isosceles, you either need to show two sides are equal or two angles are equal. Look for congruent triangles or properties of isosceles triangles.

Very Short Answer Type Questions

Question 1. In figure, AB = AC, CD = CA and ∠ADC = 20°, find ∠ABC.
Answer: In ΔADC, since CD = CA, the angles opposite to these sides are equal: ∠CDA = ∠CAD. Since ∠ADC = 20°, then ∠CAD = 20°. Now, consider ΔABC. We are given AB = AC, which means it is an isosceles triangle. Therefore, the angles opposite these equal sides are also equal: ∠ABC = ∠ACB. The exterior angle ∠BCD (which is ∠BCA + ∠ACD) would be useful if we knew ∠ACD. However, let's use the property that the sum of angles in a triangle is 180°. In ΔABC, if AB = AC, then ∠ABC = ∠ACB. From ΔADC, we have ∠ADC = 20° and ∠CAD = 20°. So, ∠ACD = 180° - 20° - 20° = 140°. This means ∠ACB = 140°. Now, in ΔABC, since ∠ABC = ∠ACB, both are 140°. However, if ∠ABC = 140° and ∠ACB = 140°, then their sum (280°) is already more than 180°, which is not possible for a triangle. This suggests an issue with the diagram or question interpretation. Let's re-read the diagram. It shows C, D, B are collinear on the base. A is the vertex. The image shows ∠ADC is 20°. In ΔADC, CD = CA. So, ∠CAD = ∠CDA = 20°. Then, ∠ACD = 180° - (20° + 20°) = 140°. This angle ∠ACD is part of ∠ACB if B is on the other side. Let's assume the question means ∠ACD is 140°. But AB = AC is given. In ΔABC, if ∠C = 140°, then ∠B must also be 140° (since AB=AC). This is impossible. There must be a misunderstanding of the figure or problem statement. Let's assume D is a point on the line segment BC. Let's use the provided solution format from the OCR (though it was incomplete): "But AB also equal to AC ⇒ ∠ABC = ∠ACB = 40°". This implies ∠A = 100°. If ∠ABC = 40° and ∠ACB = 40°, then ∠BAC = 180 - (40+40) = 100°. If ∠ADC = 20°, and CD=CA, then ∠CAD=20°. Then ∠ACD = 180 - 20 - 20 = 140°. This is not 40°. The information provided in the question `CD = CA` and `∠ADC = 20°` makes `∠ACD = 140°`. If `AB = AC`, then `∠ABC = ∠ACB`. If `∠ACB = 140°`, `∠ABC` also would be `140°`, which is incorrect. Perhaps the diagram implies D is on an extension of BC, or the original solution for Q1 (which gives 40°) is for a different problem. I must follow the source's logic if available. The source provides `∠ABC = ∠ACB = 40°`. Let's assume `∠ACD` as shown in the diagram is `∠C` of triangle ABC. If `AB = AC`, then `∠ABC = ∠ACB`. If `CD = CA`, then `ΔCDA` is isosceles. Given `∠ADC = 20°`, then `∠CAD = 20°`. The exterior angle `∠BCA` (of `ΔCDA`) = `∠CAD + ∠CDA` = `20° + 20° = 40°`. So, `∠ACB = 40°`. Since `AB = AC`, then `∠ABC = ∠ACB = 40°`. This path makes sense and gives `40°`, which is consistent with the (incomplete) original solution.
Answer: First, consider triangle ADC. We are given that CD = CA. This means that the angles opposite to these equal sides are also equal, so ∠CAD = ∠CDA. Since ∠ADC is given as 20°, then ∠CAD is also 20°. Now, look at angle ∠ACB. This angle is an exterior angle to triangle ADC at vertex C (if B, C, D are collinear, and C is between B and D). An exterior angle is equal to the sum of the two opposite interior angles. So, ∠ACB = ∠CAD + ∠CDA = 20° + 20° = 40°. Finally, consider triangle ABC. We are given that AB = AC. This means triangle ABC is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are equal. Since ∠ACB = 40°, then ∠ABC must also be 40°. Thus, ∠ABC = 40°.In simple words: Because CD equals CA, the angles opposite them in triangle ADC are equal, making ∠CAD 20°. Angle ∠ACB is an outside angle for triangle ADC, so it's the sum of ∠CAD and ∠CDA, which is 20° + 20° = 40°. Since AB equals AC in triangle ABC, the angles opposite them, ∠ABC and ∠ACB, must also be equal. So, if ∠ACB is 40°, then ∠ABC is also 40°.

🎯 Exam Tip: Remember the properties of isosceles triangles (angles opposite equal sides are equal) and the exterior angle theorem (exterior angle equals sum of opposite interior angles).

Question 2. In Δ's ABC and DEF, if AB = DF, BC = DE, AC = EF and ∠D = 55°. Find ∠B.
Answer: We are given the following information about two triangles, ΔABC and ΔDEF: 1. AB = DF 2. BC = DE 3. AC = EF These three conditions (Side-Side-Side or SSS congruence criterion) tell us that triangle ABC is congruent to triangle DFE (not DEF, because the correspondence of sides is AB=DF, BC=DE, AC=FE). Therefore, by Corresponding Parts of Congruent Triangles (c.p.c.t), all corresponding angles are equal. So, ∠A = ∠D, ∠B = ∠F, and ∠C = ∠E. We are given that ∠D = 55°. Since ∠A = ∠D, then ∠A = 55°. If the congruence is ΔABC ≅ ΔDFE, then ∠B corresponds to ∠F. If the congruence is ΔABC ≅ ΔFDE, then ∠B corresponds to ∠D. The given sides are: AB = DF, BC = DE, AC = EF. To establish correct congruence, let's map the vertices: A ↔ D (opposite to BC and DE, respectively) B ↔ F (opposite to AC and EF, respectively) C ↔ E (opposite to AB and DF, respectively) So, ΔABC ≅ ΔDFE. Therefore, the corresponding angles are: ∠A = ∠D ∠B = ∠F ∠C = ∠E We are given ∠D = 55°. The question asks for ∠B. Based on this congruence, ∠B = ∠F. We don't have ∠F. However, the source solution states `∆ABC = ∆DEF (by c.p.c.t) ∠D = ∠B = 55°`. This implies that the congruence used by the source is ΔABC ≅ ΔDEF. Let's check if this is possible with the given side equalities. If ΔABC ≅ ΔDEF, then: AB = DE (Given: AB=DF, so this is a mismatch if we follow the source's congruence statement) BC = EF (Given: BC=DE) AC = DF (Given: AC=EF) So, the source's congruence statement `∆ABC = ∆DEF` is inconsistent with the given side equalities (AB=DF, BC=DE, AC=EF). If we strictly follow the side equalities: AB=DF, BC=DE, AC=EF. This means ΔABC is congruent to ΔFDE (or ΔDFE, depends on order). A (opposite BC) ↔ F (opposite DE) B (opposite AC) ↔ D (opposite EF) C (opposite AB) ↔ E (opposite DF) So, ΔABC ≅ ΔFDE. Then ∠B (opposite AC) must be equal to ∠D (opposite EF). Thus, ∠B = ∠D. Since ∠D = 55°, then ∠B = 55°. This matches the source's final answer, even if their congruence statement was slightly miswritten. The correct correspondence for ∠B to equal ∠D is ΔABC ≅ ΔFDE.In simple words: We have two triangles, ABC and DEF. All three sides of triangle ABC are equal to the three sides of triangle FDE (AB=DF, BC=DE, AC=EF). This means the two triangles are exactly the same size and shape, which we call congruent. When triangles are congruent, their matching angles are also equal. Angle B in triangle ABC matches angle D in triangle FDE. Since angle D is 55°, angle B must also be 55°. This is a key part of triangle congruence rules.

🎯 Exam Tip: Pay close attention to the order of vertices when writing congruence statements. The sides and angles must correspond correctly (e.g., AB corresponds to DF, not DE if AB=DF is given). SSS congruence means all three sides are equal, making all corresponding angles equal.

Question 3. In figure, ∠B = ∠D = 90° and BC = CD. Is AB = DE? Why?
Answer: We need to determine if AB = DE, based on the given information and figure. Consider the two right-angled triangles, ΔABC and ΔCDE. We are given: 1. ∠B = ∠D = 90° (Both are right angles) 2. BC = CD (A pair of corresponding sides are equal) 3. ∠ACB = ∠DCE (These are vertically opposite angles, formed when two straight lines intersect, and vertically opposite angles are always equal). Now, we have two angles and one included side equal: - Angle ∠B = Angle ∠D (Both 90°) - Side BC = Side CD (Given) - Angle ∠ACB = Angle ∠DCE (Vertically opposite angles) This fulfills the Angle-Side-Angle (ASA) congruence property. Therefore, ΔABC is congruent to ΔCDE (ΔABC ≅ ΔCDE). Since the triangles are congruent, their corresponding parts are equal (c.p.c.t). Thus, the side AB corresponds to the side DE. Therefore, AB = DE. Yes, AB and DE are equal.In simple words: We have two triangles, ABC and CDE. Both have a 90-degree angle. The side BC is equal to side CD. Also, the angles ∠ACB and ∠DCE are vertically opposite, so they are equal. Because of these three facts (Angle-Side-Angle), the two triangles are exactly the same size and shape. Since they are identical, their matching sides must also be equal, so AB is equal to DE.

🎯 Exam Tip: The ASA congruence rule is powerful: if two angles and the *included* side of one triangle are equal to two angles and the *included* side of another triangle, then the triangles are congruent. Always look for vertically opposite angles or common sides when proving congruence.

Question 5. In figure, AB = 7 cm, BC = 8 cm and AC = 7.6 cm then write the (i) greatest angle of the triangle (ii) smallest angle of the triangle
Answer: In any triangle, the angle opposite the longest side is the greatest angle, and the angle opposite the shortest side is the smallest angle. First, let's identify the side lengths: AB = 7 cm BC = 8 cm AC = 7.6 cm From these lengths, we can see: - The longest side is BC = 8 cm. - The shortest side is AB = 7 cm. (i) The greatest angle is the angle opposite the longest side (BC). The angle opposite to BC is ∠CAB (or simply ∠A). So, the greatest angle is ∠CAB. (ii) The smallest angle is the angle opposite the shortest side (AB). The angle opposite to AB is ∠ACB (or simply ∠C). So, the smallest angle is ∠ACB.In simple words: In any triangle, the biggest angle is always across from the longest side, and the smallest angle is always across from the shortest side. Here, the longest side is BC, so the angle opposite it (∠CAB) is the greatest. The shortest side is AB, so the angle opposite it (∠ACB) is the smallest. This rule helps us understand the relationship between side lengths and angles.

🎯 Exam Tip: Always remember the fundamental triangle property: the side opposite the largest angle is the longest side, and the side opposite the smallest angle is the shortest side. This is crucial for comparing angles and sides.

Question 6. In ΔPQR, ∠Q = 35°, ∠R = 61° and the bisector of ∠QPR meet QR at x. Then arrange the sides PX, QX and RX in descending order of their length.
Answer: First, let's find the third angle of triangle PQR, which is ∠P (or ∠QPR). ∠QPR + ∠Q + ∠R = 180° (Sum of angles in a triangle). ∠QPR + 35° + 61° = 180°. ∠QPR + 96° = 180°. ∠QPR = 180° - 96° = 84°. Now, PX is the bisector of ∠QPR, so it divides ∠QPR into two equal halves. ∠QPX = ∠RPX = ∠QPR / 2 = 84° / 2 = 42°. Next, let's find the angles in ΔPQX and ΔPRX. In ΔPQX: ∠Q = 35° ∠QPX = 42° ∠PXQ = 180° - (35° + 42°) = 180° - 77° = 103°. In ΔPRX: ∠R = 61° ∠RPX = 42° ∠PXR = 180° - (61° + 42°) = 180° - 103° = 77°. (Also, ∠PXQ and ∠PXR are supplementary, 103° + 77° = 180°, which confirms our calculation). Now, to arrange the sides QX, PX, RX in descending order, we need to compare the angles opposite to them. Consider ΔPQX: Sides: QX, PX, PQ Angles: ∠QPX (42°), ∠PQX (35°), ∠PXQ (103°) Opposite QX is ∠QPX = 42°. Opposite PX is ∠PQX = 35°. Opposite PQ is ∠PXQ = 103°. So, in ΔPQX, the order of sides (descending) is PQ > QX > PX. Consider ΔPRX: Sides: RX, PX, PR Angles: ∠RPX (42°), ∠PRX (61°), ∠PXR (77°) Opposite RX is ∠RPX = 42°. Opposite PX is ∠PRX = 61°. Opposite PR is ∠PXR = 77°. So, in ΔPRX, the order of sides (descending) is PR > PX > RX. The question asks to arrange QX, PX, and RX in descending order. From ΔPQX: QX (opposite 42°) and PX (opposite 35°). So, QX > PX. From ΔPRX: PX (opposite 61°) and RX (opposite 42°). So, PX > RX. Combining these, we get QX > PX > RX.In simple words: First, find all the angles in the main triangle PQR and then in the smaller triangles created by the angle bisector PX. In a triangle, the side opposite a larger angle is longer. By comparing the angles in triangles PQX and PRX, we find that the side QX is longer than PX, and PX is longer than RX. This means the order from longest to shortest is QX, then PX, then RX.

🎯 Exam Tip: To compare side lengths in a triangle, first find all interior angles. The side opposite the largest angle is the longest, and the side opposite the smallest angle is the shortest.

Short Answer Type Questions

Question 2. In figure, BA ⊥ AC, DE ⊥ EF, such that BA = DE and BF = CD, prove that AC = EF.
Answer: We need to prove that AC = EF. To do this, we can show that ΔABC is congruent to ΔDEF. Let's analyze the given information: 1. BA ⊥ AC, which means ∠BAC = 90°. 2. DE ⊥ EF, which means ∠DEF = 90°. So, we have ∠BAC = ∠DEF = 90° (both are right angles). 3. We are given BA = DE. This is one pair of equal sides. 4. We are given BF = CD. Let's add FC to both sides of the equation BF = CD. BF + FC = CD + FC This simplifies to BC = FD. Now, consider ΔABC and ΔDEF: - ∠BAC = ∠DEF (Both are 90°, as shown above) - BA = DE (Given) - BC = FD (Proved above) With a right angle, hypotenuse, and one side equal, we can use the RHS (Right angle - Hypotenuse - Side) congruence property. Therefore, ΔABC is congruent to ΔDEF (ΔABC ≅ ΔDEF) by RHS congruence. Since the triangles are congruent, their corresponding parts are equal (c.p.c.t). Thus, AC = EF. Hence proved.In simple words: We want to show that side AC is the same length as side EF. We are given that both triangles (ABC and DEF) have a right angle. We know that side BA is equal to side DE. Also, by adding a common part (FC) to the given equal lengths BF and CD, we find that side BC is equal to side FD. Since both triangles have a right angle, their hypotenuses (BC and FD) are equal, and one other side (BA and DE) is equal, the two triangles are congruent by the RHS rule. Because they are congruent, all their matching sides are equal, so AC must be equal to EF.

🎯 Exam Tip: When dealing with right-angled triangles, always check for the RHS congruence criterion (Right angle, Hypotenuse, Side). Don't forget that adding a common segment to equal lengths can help establish equality of other sides.

Question 3. In figure, CB = AD and AB = CD. Can we say ∠ABC and ∠ADC are equal? Why?
Answer: To determine if ∠ABC and ∠ADC are equal, we can try to prove that the triangles containing these angles are congruent. Consider ΔABC and ΔADC. We are given: 1. CB = AD (One pair of sides) 2. AB = CD (Another pair of sides) 3. AC = AC (This is a common side to both triangles) Since all three corresponding sides of ΔABC and ΔADC are equal (AB=CD, BC=AD, AC=AC), these two triangles are congruent by the SSS (Side-Side-Side) congruence property. Therefore, ΔABC ≅ ΔADC. When two triangles are congruent, their corresponding parts are equal (c.p.c.t). The angle ∠ABC is opposite the side AC in ΔABC. The angle ∠ADC is opposite the side AC in ΔADC. Since AC is the common side, the angles opposite to it in congruent triangles must be equal. Thus, ∠ABC = ∠ADC. Yes, we can say that ∠ABC and ∠ADC are equal because the triangles ΔABC and ΔADC are congruent by the SSS property.In simple words: We have two triangles, ABC and ADC. We are given that side CB is equal to side AD, and side AB is equal to side CD. Also, the side AC is common to both triangles. Since all three sides of triangle ABC are equal to the three sides of triangle ADC, the triangles are exactly the same. Because they are congruent, their matching angles are also equal. Angle ABC matches angle ADC, so they are equal.

🎯 Exam Tip: The SSS (Side-Side-Side) congruence criterion is very direct: if all three sides of one triangle are equal to all three sides of another triangle, they are congruent. This immediately means all their corresponding angles are also equal.

Question 4. In figure, CN ⊥ AB, DM ⊥ AB and CN = DM. Are OC and OD equal? Why?
Answer: We need to determine if OC and OD are equal. We can do this by proving that triangles ΔONC and ΔOMD are congruent. Let's analyze the given information: 1. CN ⊥ AB, which means ∠ONC = 90°. 2. DM ⊥ AB, which means ∠OMD = 90°. So, we have ∠ONC = ∠OMD = 90° (both are right angles). 3. We are given CN = DM (a pair of corresponding sides are equal). 4. Observe the angles ∠NOC and ∠MOD. These are vertically opposite angles, formed by the intersection of lines CD and MN. Vertically opposite angles are always equal. So, ∠NOC = ∠MOD. Now, consider ΔONC and ΔOMD: - Angle ∠ONC = Angle ∠OMD (Both 90°) - Angle ∠NOC = Angle ∠MOD (Vertically opposite angles) - Side CN = Side DM (Given) This fulfills the AAS (Angle-Angle-Side) congruence property (two angles and a non-included side are equal). Therefore, ΔONC is congruent to ΔOMD (ΔONC ≅ ΔOMD). Since the triangles are congruent, their corresponding parts are equal (c.p.c.t). The side OC in ΔONC corresponds to the side OD in ΔOMD. Thus, OC = OD. Yes, OC and OD are equal.In simple words: We have two triangles, ONC and OMD. Both have a 90-degree angle because CN and DM are perpendicular to AB. The side CN is equal to the side DM. Also, the angles at O (∠NOC and ∠MOD) are vertically opposite, so they are equal. Since two angles and one non-included side of triangle ONC are equal to the corresponding parts of triangle OMD, the triangles are congruent (exact copies). This means their matching parts, OC and OD, are also equal.

🎯 Exam Tip: Look for vertical angles and perpendicular lines. The AAS congruence criterion is very useful when the equal side is not between the two equal angles.

Question 5. In figure, ∠A = 35°, ∠ABC = 100° and BD ⊥ AC, prove that ABDC is an isosceles triangle.
Answer: We need to prove that triangle BDC is an isosceles triangle. For a triangle to be isosceles, two of its sides must be equal, or two of its angles must be equal. Let's find the angles in triangle ABC first. In ΔABC, the sum of angles is 180°. ∠A + ∠ABC + ∠ACB = 180°. 35° + 100° + ∠ACB = 180°. 135° + ∠ACB = 180°. ∠ACB = 180° - 135° = 45°. So, ∠C (which is ∠DCB in ΔBDC) = 45°. Next, consider ΔABD. We are given BD ⊥ AC, which means ∠ADB = 90°. In ΔABD, the sum of angles is 180°. ∠A + ∠ABD + ∠ADB = 180°. 35° + ∠ABD + 90° = 180°. 125° + ∠ABD = 180°. ∠ABD = 180° - 125° = 55°. Now, we need to find ∠DBC. We know ∠ABC = 100° and ∠ABD = 55°. So, ∠DBC = ∠ABC - ∠ABD = 100° - 55° = 45°. Now, let's look at triangle BDC. We have found: - ∠DCB = 45° - ∠DBC = 45° Since two angles of ΔBDC are equal (∠DCB = ∠DBC = 45°), the sides opposite to these angles must also be equal. The side opposite ∠DCB is BD. The side opposite ∠DBC is CD. Therefore, BD = CD. Since two sides of ΔBDC are equal, ΔBDC is an isosceles triangle. Hence proved.In simple words: To prove that triangle BDC is isosceles, we need to show that two of its angles or two of its sides are equal. First, we find angle C in triangle ABC, which is 45°. Then, we find angle ABD in triangle ABD, which is 55°. Using these, we can find angle DBC, which is 100° - 55° = 45°. So, in triangle BDC, we have two angles equal to 45° (∠DCB and ∠DBC). When a triangle has two equal angles, the sides opposite those angles are also equal. This makes triangle BDC an isosceles triangle.

🎯 Exam Tip: When proving a triangle is isosceles, a common strategy is to find two equal angles. Remember to use the sum of angles in a triangle (180°) and properties of perpendicular lines to find unknown angles.

Question 6. In the given figure, O is the middle point of both AB and CD. Prove that AC = BD and AC || BD.
Answer: We are given that O is the midpoint of both AB and CD. This means: 1. OA = OB (since O is the midpoint of AB) 2. OC = OD (since O is the midpoint of CD) We need to prove that AC = BD and AC || BD. Consider ΔAOC and ΔBOD. We have: - OA = OB (Proved above) - OC = OD (Proved above) - ∠AOC = ∠BOD (These are vertically opposite angles, and vertically opposite angles are always equal). Now, we have two sides and the included angle equal (SAS congruence property). Therefore, ΔAOC is congruent to ΔBOD (ΔAOC ≅ ΔBOD). Since the triangles are congruent, their corresponding parts are equal (c.p.c.t). 1. The side AC in ΔAOC corresponds to the side BD in ΔBOD. So, AC = BD. (First part of the proof) 2. Also, corresponding angles are equal. So, ∠OAC = ∠OBD. These angles (∠OAC and ∠OBD) are alternate interior angles for lines AC and BD with transversal AB. Since the alternate interior angles are equal, the lines AC and BD must be parallel. Therefore, AC || BD. (Second part of the proof) Hence proved.In simple words: We are told that O is the middle point for both lines AB and CD. This means OA=OB and OC=OD. If we look at triangles AOC and BOD, we see that two sides (OA and OC; OB and OD) are equal, and the angles between them (∠AOC and ∠BOD) are also equal because they are vertically opposite. So, these two triangles are exactly the same (congruent by SAS rule). Because they are congruent, their matching parts are equal, which means side AC is equal to side BD. Also, their matching angles, ∠OAC and ∠OBD, are equal. Since these are alternate interior angles, it means that line AC must be parallel to line BD.

🎯 Exam Tip: Look for midpoints as they provide equal segments, and vertically opposite angles as they provide equal angles. SAS congruence is a powerful tool to prove triangles are congruent, which then allows you to prove equality of other sides and angles.

Question 8. In the following figures, two sides of ∆АВС, AB and BC and median AD are respectively equal to PQ, QR and median PM of ∆PQR. Prove that: ∆АВC = ∆PQR.
Answer: We are given the following information: 1. AB = PQ (given) 2. BC = QR (given) 3. Median AD of ΔABC is equal to median PM of ΔPQR (AD = PM). We need to prove that ΔABC is congruent to ΔPQR. Since AD is the median to BC, D is the midpoint of BC. So, BD = DC = (1/2)BC. Similarly, since PM is the median to QR, M is the midpoint of QR. So, QM = MR = (1/2)QR. Since BC = QR (given), then (1/2)BC = (1/2)QR, which means BD = QM. Now, consider ΔABD and ΔPQM: - AB = PQ (given) - AD = PM (given) - BD = QM (proved above) By SSS (Side-Side-Side) congruence criterion, ΔABD ≅ ΔPQM. Since these triangles are congruent, their corresponding parts are equal (c.p.c.t). Therefore, ∠B = ∠Q. Now, consider ΔABC and ΔPQR: - AB = PQ (given) - ∠B = ∠Q (proved above) - BC = QR (given) By SAS (Side-Angle-Side) congruence criterion, ΔABC ≅ ΔPQR. Hence proved.In simple words: We want to show that triangle ABC is exactly the same as triangle PQR. We are given that sides AB and BC of triangle ABC are equal to sides PQ and QR of triangle PQR. We are also given that the median AD is equal to the median PM. Medians divide the opposite side into two equal halves. So, we first prove that the smaller triangles ABD and PQM are congruent using SSS. From this, we know that angle B is equal to angle Q. Now, using sides AB=PQ, angle B=angle Q, and sides BC=QR, we can say that triangle ABC is congruent to triangle PQR by the SAS rule.

🎯 Exam Tip: When medians are involved, remember they divide the opposite side into two equal halves. This often helps establish side equalities for proving congruence of smaller triangles first, which then helps prove congruence of the larger triangles.

Question 9. In figure, side QR of ΔPQR is produced on both sides such that ∠PQS = ∠PRT. Prove that PQ = PR.
Answer: We need to prove that PQ = PR. For this, we can show that ∠PQR = ∠PRQ, as sides opposite to equal angles in a triangle are equal. Let's use the property of linear pairs. Angles ∠PQS and ∠PQR form a linear pair on the straight line S-Q-R. So, ∠PQS + ∠PQR = 180° (linear pair angles). This means ∠PQR = 180° - ∠PQS. (Equation 1) Similarly, angles ∠PRQ and ∠PRT form a linear pair on the straight line Q-R-T. So, ∠PRQ + ∠PRT = 180° (linear pair angles). This means ∠PRQ = 180° - ∠PRT. (Equation 2) We are given that ∠PQS = ∠PRT. (Equation 3) Now, substitute Equation 3 into Equation 1: ∠PQR = 180° - ∠PRT. (Equation 4) Comparing Equation 2 and Equation 4, we can see that both ∠PQR and ∠PRQ are equal to (180° - ∠PRT). Therefore, ∠PQR = ∠PRQ. In ΔPQR, since two angles (∠PQR and ∠PRQ) are equal, the sides opposite to these angles must also be equal. The side opposite ∠PQR is PR. The side opposite ∠PRQ is PQ. Thus, PR = PQ. Hence proved.In simple words: We want to show that side PQ is equal to side PR. We start by using the fact that angles on a straight line add up to 180 degrees. So, angle PQR can be found from angle PQS, and angle PRQ can be found from angle PRT. Since we are given that angle PQS equals angle PRT, it means angle PQR must also equal angle PRQ. When two angles in a triangle are equal, the sides opposite those angles are also equal. Therefore, PQ equals PR.

🎯 Exam Tip: Remember that angles forming a linear pair add up to 180°. The converse of the isosceles triangle theorem (sides opposite equal angles are equal) is crucial for proving equality of sides when angles are known to be equal.

Question 10. Prove that the medians bisecting the equal sides of an isosceles triangle are equal.
Answer: Let's consider an isosceles triangle ABC, where AB = AC. Let D be the midpoint of AB, so CD is the median bisecting AB. Let E be the midpoint of AC, so BE is the median bisecting AC. We need to prove that these medians are equal, i.e., BE = CD. Here's the proof: 1. Since ΔABC is an isosceles triangle with AB = AC (given). (Equation 1) 2. In an isosceles triangle, the angles opposite the equal sides are also equal. So, ∠ABC = ∠ACB. (Equation 2) 3. D is the midpoint of AB, so AD = DB = (1/2)AB. 4. E is the midpoint of AC, so AE = EC = (1/2)AC. Since AB = AC (from Equation 1), then (1/2)AB = (1/2)AC. This means DB = EC and AD = AE. (Equation 3) Now, consider ΔBCD and ΔCBE. - BC = CB (This is a common side to both triangles). - ∠DBC = ∠ECB (This is ∠ABC = ∠ACB from Equation 2). - DB = EC (From Equation 3). By SAS (Side-Angle-Side) congruence property, ΔBCD ≅ ΔCBE. Since the triangles are congruent, their corresponding parts are equal (c.p.c.t). The side CD in ΔBCD corresponds to the side BE in ΔCBE. Therefore, CD = BE. Hence proved.In simple words: Imagine an isosceles triangle ABC where sides AB and AC are equal. Let's draw a line from corner C to the middle of side AB (called median CD), and another line from corner B to the middle of side AC (called median BE). We want to show that these two medians (CD and BE) are the same length. Since AB=AC, then half of AB equals half of AC, so DB=EC. Also, in an isosceles triangle, the angles at the base (∠ABC and ∠ACB) are equal. Now, if we look at two smaller triangles, BCD and CBE, they share side BC, have equal angles at the base (∠DBC and ∠ECB), and equal sides (DB and EC). This means they are congruent by the SAS rule. Because they are congruent, their third sides (CD and BE) must also be equal.

🎯 Exam Tip: When proving medians are equal in an isosceles triangle, use the SAS congruence criterion by considering the two triangles formed by the medians and the base. The common base and equal base angles are key elements.

Long Answer Type Questions

Question 1. "If two sides of a triangle are unequal then the longer side has greater angle opposite to it." Prove it.
Answer: To prove the theorem: "If two sides of a triangle are unequal, the angle opposite the longer side is greater." Let's consider a triangle ABC where AC > AB. We need to prove that ∠ABC > ∠ACB. Construction: Mark a point D on side AC such that AB = AD. This is possible because AC is longer than AB. Join B to D, forming line segment BD. Proof: 1. In ΔABD, we have AB = AD (by construction). This means ΔABD is an isosceles triangle. Therefore, the angles opposite the equal sides are equal: ∠ABD = ∠ADB. (Equation 1) 2. Now, consider ΔBCD. ∠ADB is an exterior angle to ΔBCD. According to the exterior angle theorem, an exterior angle of a triangle is equal to the sum of its two opposite interior angles. So, ∠ADB = ∠DBC + ∠DCB (which is ∠ACB). This means ∠ADB > ∠DCB (or ∠ACB). (Equation 2) 3. From Equation 1, we know ∠ABD = ∠ADB. Substituting this into Equation 2, we get ∠ABD > ∠ACB. (Equation 3) 4. Now, observe the angle ∠ABC. ∠ABC is composed of two parts: ∠ABD + ∠DBC. So, ∠ABC > ∠ABD. (Equation 4) 5. From Equation 3, we have ∠ABD > ∠ACB. From Equation 4, we have ∠ABC > ∠ABD. Combining these inequalities, if ∠ABC is greater than ∠ABD, and ∠ABD is greater than ∠ACB, then it logically follows that ∠ABC is greater than ∠ACB. Therefore, ∠ABC > ∠ACB. Hence proved.In simple words: We want to show that if one side of a triangle is longer than another side, the angle opposite the longer side is bigger. We start with a triangle ABC where AC is longer than AB. We mark a point D on AC so that AB and AD are equal. This creates a small isosceles triangle ABD, where angles ∠ABD and ∠ADB are equal. Angle ∠ADB is an outside angle for the smaller triangle BDC, so it must be bigger than angle C. Since angle ∠ABC is bigger than angle ∠ABD, and angle ∠ABD is bigger than angle C, it means angle ∠ABC is bigger than angle C. This proves that the angle opposite the longer side (AC) is greater than the angle opposite the shorter side (AB).

🎯 Exam Tip: When proving inequalities in triangles, construction is often key. Drawing an auxiliary line to create an isosceles triangle or to define an exterior angle can simplify the proof significantly. Remember the exterior angle theorem and properties of isosceles triangles.

Question 2. In figure, PQR is a triangle and S is any point in its interior, show that SQ + SR < PQ + PR.
Answer: To prove: SQ + SR < PQ + PR. Given: S is any point in the interior of ΔPQR. Construction: Produce the line segment QS to meet PR at point T. Proof: 1. Consider ΔPQT. According to the triangle inequality theorem, the sum of any two sides of a triangle is greater than the third side. So, PQ + PT > QT. We can break QT into QS + ST. Therefore, PQ + PT > QS + ST. (Equation 1) 2. Now, consider ΔRST. Again, by the triangle inequality theorem: ST + TR > RS. (Equation 2) 3. Now, let's add Equation 1 and Equation 2: (PQ + PT) + (ST + TR) > (QS + ST) + (RS) PQ + PT + ST + TR > QS + ST + RS 4. We can subtract ST from both sides of the inequality, as it appears on both sides. PQ + PT + TR > QS + RS 5. Observe that PT + TR is equal to PR (from the figure, T lies on PR). Substitute PR for (PT + TR): PQ + PR > QS + RS. Rearranging this inequality to match the required format: SQ + SR < PQ + PR. Hence proved.In simple words: We want to show that the sum of the distances from point S to Q and S to R is less than the sum of the sides PQ and PR. We extend the line QS until it touches the side PR at point T. Now, we use the rule that in any triangle, two sides added together are always longer than the third side. First, in triangle PQT, PQ + PT is longer than QT (which is QS + ST). Second, in triangle RST, ST + TR is longer than RS. By adding these two inequalities together and simplifying, we get that PQ + PR is longer than QS + SR, which is what we wanted to prove. This illustrates how distances within a triangle relate to its outer boundaries.

🎯 Exam Tip: The triangle inequality theorem (sum of any two sides is greater than the third side) is fundamental for proving inequalities. When a point is inside a polygon, drawing a line from the point to a vertex and extending it to intersect an opposite side often helps create new triangles for applying the theorem.

Question 4. In figure, side AB and AC are produced (RBSESolutions.com) to point D and E respectively. Bisectors BO and CO of ∠DBC and ∠ECB respectively meet at O. If AB > AC. Prove that OC > OB.
Answer: We are given a triangle \(ABC\). Sides \(AB\) and \(AC\) are extended to points \(D\) and \(E\), creating exterior angles. The lines \(BO\) and \(CO\) divide these exterior angles (\(\angle DBC\) and \(\angle ECB\)) into two equal parts. We also know that side \(AB\) is longer than side \(AC\) (\(AB > AC\)).
In triangle \(ABC\), because \(AB > AC\), the angle opposite to \(AB\) (\(\angle ACB\)) must be greater than the angle opposite to \(AC\) (\(\angle ABC\)). So, \( \angle ACB > \angle ABC \). This is a fundamental property of triangles relating side lengths to opposite angles.
We know that \(\angle DBC = 180^\circ - \angle ABC\) and \(\angle ECB = 180^\circ - \angle ACB\) because they form linear pairs.
Since \(BO\) bisects \(\angle DBC\), we have \( \angle OBC = \frac{1}{2} \angle DBC = \frac{1}{2} (180^\circ - \angle ABC) = 90^\circ - \frac{1}{2} \angle ABC \).
Since \(CO\) bisects \(\angle ECB\), we have \( \angle OCB = \frac{1}{2} \angle ECB = \frac{1}{2} (180^\circ - \angle ACB) = 90^\circ - \frac{1}{2} \angle ACB \).
Because \( \angle ACB > \angle ABC \), if we multiply both sides by \( -\frac{1}{2} \), the inequality reverses: \( -\frac{1}{2} \angle ACB < -\frac{1}{2} \angle ABC \).
Now, adding \(90^\circ\) to both sides, we get: \( 90^\circ - \frac{1}{2} \angle ACB < 90^\circ - \frac{1}{2} \angle ABC \).
This means \( \angle OCB < \angle OBC \).
In triangle \(OBC\), the side opposite to the smaller angle is smaller. Since \(\angle OCB < \angle OBC\), the side opposite \(\angle OCB\) (which is \(OB\)) must be smaller than the side opposite \(\angle OBC\) (which is \(OC\)).
Therefore, \(OB < OC\), or \(OC > OB\). This proves the statement.
In simple words: When one side of a triangle is longer, the angle opposite to it is bigger. Using this, and how bisectors split the angles, we can show that if \(AB\) is longer than \(AC\), then the distance \(OC\) will be longer than \(OB\).

🎯 Exam Tip: When proving inequalities involving sides and angles, always remember that the angle opposite the longer side is greater, and the side opposite the greater angle is longer. Use linear pair properties for exterior angles.

Question 5. If figure, ABCD is a quadrilateral. Prove that AB + BC + CD + DA > AC + BD.
Answer: We need to prove that the sum of all sides of a quadrilateral is greater than the sum of its diagonals. This proof relies on the Triangle Inequality Theorem, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This theorem is a fundamental concept in geometry, ensuring that a triangle can actually be formed with given side lengths.
Consider the quadrilateral \(ABCD\) with diagonals \(AC\) and \(BD\).
In triangle \(ABC\), according to the triangle inequality:
\(AB + BC > AC\) ...(i)
In triangle \(ADC\), according to the triangle inequality:
\(AD + DC > AC\) ...(ii)
In triangle \(ABD\), according to the triangle inequality:
\(AB + AD > BD\) ...(iii)
In triangle \(BCD\), according to the triangle inequality:
\(BC + CD > BD\) ...(iv)
Now, we add all four inequalities (i), (ii), (iii), and (iv) together:
\((AB + BC) + (AD + DC) + (AB + AD) + (BC + CD) > AC + AC + BD + BD\)
Combine like terms on both sides:
\(2AB + 2BC + 2CD + 2DA > 2AC + 2BD\)
Finally, divide the entire inequality by 2:
\(AB + BC + CD + DA > AC + BD\)
This proves that the sum of the lengths of the sides of any quadrilateral is greater than the sum of the lengths of its diagonals.
In simple words: Imagine a four-sided shape. If you add up the lengths of all its outer edges, that total will always be bigger than if you add up the lengths of its two inside diagonal lines. We prove this by using the rule that any two sides of a triangle are always longer than its third side, applying it to the four small triangles formed by the diagonals.

🎯 Exam Tip: This is a standard proof using the triangle inequality theorem. Make sure to apply the theorem to all four triangles formed by the quadrilateral's sides and diagonals correctly.

Question 6. In the given figure, PQRS is a quadrilateral. PQ is its (RBSESolutions.com) longest side and RS is its shortest side. Prove that: ∠R > ∠P and ∠S > ∠Q.
Answer: We need to prove that certain angles in the quadrilateral are larger than others, based on the lengths of its sides. We are given a quadrilateral \(PQRS\) where \(PQ\) is the longest side and \(RS\) is the shortest side. The key principle here is that in any triangle, the angle opposite a longer side is greater than the angle opposite a shorter side.
To prove this, we first draw the diagonals \(PR\) and \(QS\), which divide the quadrilateral into smaller triangles.
**Part 1: Proving \( \angle R > \angle P \)**
Consider triangle \(PQR\).
We know \(PQ\) is the longest side of the quadrilateral, so \(PQ > QR\).
Applying the angle-side relationship in \(\Delta PQR\): the angle opposite \(PQ\) (\(\angle QRP\)) is greater than the angle opposite \(QR\) (\(\angle QPR\)).
\( \implies \angle QRP > \angle QPR \) ...(i)
Next, consider triangle \(PSR\).
We know \(RS\) is the shortest side of the quadrilateral, so \(PS > RS\).
Applying the angle-side relationship in \(\Delta PSR\): the angle opposite \(PS\) (\(\angle SRP\)) is greater than the angle opposite \(RS\) (\(\angle SPR\)).
\( \implies \angle SRP > \angle SPR \) ...(ii)
Now, add inequalities (i) and (ii):
\( (\angle QRP + \angle SRP) > (\angle QPR + \angle SPR) \)
The sum \( \angle QRP + \angle SRP \) forms the entire angle \(\angle R\) of the quadrilateral.
The sum \( \angle QPR + \angle SPR \) forms the entire angle \(\angle P\) of the quadrilateral.
Thus, \( \angle R > \angle P \).
**Part 2: Proving \( \angle S > \angle Q \)**
Consider triangle \(PQS\).
Since \(PQ\) is the longest side of the quadrilateral, \(PQ > PS\).
Applying the angle-side relationship in \(\Delta PQS\): the angle opposite \(PQ\) (\(\angle PSQ\)) is greater than the angle opposite \(PS\) (\(\angle PQS\)).
\( \implies \angle PSQ > \angle PQS \) ...(iii)
Next, consider triangle \(QRS\).
Since \(RS\) is the shortest side of the quadrilateral, \(QR > RS\).
Applying the angle-side relationship in \(\Delta QRS\): the angle opposite \(QR\) (\(\angle RSQ\)) is greater than the angle opposite \(RS\) (\(\angle RQS\)).
\( \implies \angle RSQ > \angle RQS \) ...(iv)
Now, add inequalities (iii) and (iv):
\( (\angle PSQ + \angle RSQ) > (\angle PQS + \angle RQS) \)
The sum \( \angle PSQ + \angle RSQ \) forms the entire angle \(\angle S\) of the quadrilateral.
The sum \( \angle PQS + \angle RQS \) forms the entire angle \(\angle Q\) of the quadrilateral.
Thus, \( \angle S > \angle Q \).
Both parts of the statement are now proven. This demonstrates how side lengths influence the magnitude of angles in a geometric figure.
In simple words: In any triangle, the angle facing the longest side is always the biggest. If we draw lines inside the quadrilateral from corner to corner, we make four small triangles. By using this rule in each of those triangles, we can show that the angles opposite the very longest side of the quadrilateral (\(\angle R\)) and the angle opposite the very shortest side (\(\angle S\)) are larger than their opposite corner angles (\(\angle P\) and \(\angle Q\)) respectively.

🎯 Exam Tip: This type of proof requires dividing the quadrilateral into triangles using its diagonals. Clearly state which triangle you are considering and which sides/angles are involved in each step of the inequality.