RS Aggarwal Class 9 Mathematics Solutions Chapter 9 Quadrilaterals and Parallelograms

Exercise 9A

Question 1. Three angles of a quadrilateral are 56°, 115° and 84°. Find the fourth angle.
Answer: Let the fourth angle be x. The sum of angles in any quadrilateral equals 360°.

56° + 115° + 84° + x = 360°

255° + x = 360°

x = 360° - 255° = 105°

The fourth angle is 105°.
In simple words: Add up the three given angles. Then subtract that total from 360° to find the missing angle.

Exam Tip: Always remember that the sum of all four angles in any quadrilateral is 360°. This is a key fact for solving angle problems.

Question 2. The angles of a quadrilateral are in the ratio 2:4:5:7. Find all four angles.
Answer: Let the angles be 2x, 4x, 5x, and 7x. Since the sum of angles in a quadrilateral is 360°:

2x + 4x + 5x + 7x = 360°

18x = 360°

x = 360°/18 = 20°

Therefore:
2x = 2 × 20 = 40°
4x = 4 × 20 = 80°
5x = 5 × 20 = 100°
7x = 7 × 20 = 140°

The four angles are 40°, 80°, 100°, and 140°.
In simple words: When angles are in a ratio, set them as multiples of a variable. Add them all up to equal 360°, then solve for that variable.

Exam Tip: Always verify your answer by adding the four angles to confirm they sum to 360°.

Question 3. Since AB ∥ DC, ∠A and ∠D are consecutive interior angles. Consecutive interior angles sum to 180°. So, ∠A + ∠D = 180° ⇒ 55° + ∠D = 180° ⇒ ∠D = 180° - 55° = 125°. Also, the sum of the angles of a quadrilateral is 360°. So, ∠A + ∠B + ∠C + ∠D = 360° ⇒ 55° + 70° + ∠C + 125° = 360° ⇒ 250° + ∠C = 360° ⇒ ∠C = 360° - 250° = 110°. Therefore, ∠C = 110° and ∠D = 125°.
Answer: Since AB ∥ DC, angles A and D are consecutive interior angles and must sum to 180°.

55° + ∠D = 180°
∠D = 125°

Using the fact that all angles in a quadrilateral sum to 360°:
55° + 70° + ∠C + 125° = 360°
250° + ∠C = 360°
∠C = 110°

Thus, ∠C = 110° and ∠D = 125°.
In simple words: When two sides are parallel, the angles on the same side add up to 180°. Use this plus the 360° rule for quadrilaterals to find the missing angles.

Exam Tip: Always identify parallel lines first - they give you a shortcut using the consecutive interior angle property.

Question 4. Given: △EDC is an equilateral triangle and △ABC is a square. To Prove: (i) AE = BE and (ii) ∠DAE = 15°
Answer: (i) Since △EDC is equilateral, all angles equal 60°. Since △ABC is a square, all angles equal 90°. In △EDA, ∠EDA = ∠EDC + ∠CDA = 60° + 90° = 150°. In △ECB, ∠ECB = ∠ECD + ∠DCB = 60° + 90° = 150°. By Side-Angle-Side criterion, △EDA ≅ △ECB. Thus, corresponding parts are equal: AE = BE.

(ii) Since ED = DA (from the equilateral and square properties) and ED = EC, we have △EDA is isosceles with ED = DA. Therefore ∠DEA = ∠DAE. In △EDA: ∠EDA + ∠DEA + ∠DAE = 180° ⇒ 150° + ∠DAE + ∠DAE = 180° ⇒ 2∠DAE = 30° ⇒ ∠DAE = 15°.
In simple words: First show the two triangles are congruent to prove the sides equal. Then use the isosceles triangle property and angle sum to find the angle measure.

Exam Tip: When a figure contains both an equilateral triangle and a square, look for congruent triangles using their known properties.

Question 5. Given: BM ⊥ AC and DN ⊥ AC and BM = DN. To Prove: AC bisects BD.
Answer: We have ∠DON = ∠MOB (vertically opposite angles) and ∠DNO = ∠BMO = 90° (given). Since BM = DN (given), by Angle-Angle-Side criterion, △DNO ≅ △BMO. Thus, corresponding parts are equal: OD = OB. Therefore, AC bisects BD.
In simple words: Both segments are perpendicular to the same line and have equal length. This makes the two triangles they form congruent, proving the bisection.

Exam Tip: Look for vertically opposite angles and equal perpendicular segments - these are strong indicators for congruent triangles.

Question 6. Given: ABCD is a quadrilateral in which AB = AD and BC = DC. To Prove: (i) AC bisects ∠A and ∠C, (ii) BE = DE, (iii) ∠ABC = ∠ADC
Answer: (i) In △ABC and △ADC, we have AB = AD (given), BC = DC (given), and AC = AC (common). By Side-Side-Side criterion, △ABC ≅ △ADC. Thus ∠BAC = ∠DAC and ∠BCA = ∠DCA. This means AC bisects ∠BAD (that is ∠A) and ∠BCD (that is ∠C).

(ii) In △ABE and △ADE, we have AB = AD (given), ∠BAE = ∠DAE (from part i), and AE = AE (common). By Side-Angle-Side criterion, △ABE ≅ △ADE. Thus BE = DE.

(iii) From the congruence △ABC ≅ △ADC established in part (i), corresponding parts are equal. Therefore, ∠ABC = ∠ADC.
In simple words: Equal sides create congruent triangles. Congruent triangles give you equal angles and equal sides, which proves all three statements.

Exam Tip: Always use congruent triangle results to find angles and sides in the rest of the figure.

Question 8. Given: O is a point within a quadrilateral ABCD. To Prove: OA + OB + OC + OD > AC + BD. Construction: Join AC and BD. Proof: In △ACO, OA + OC > AC ... (i). In △BDO, OB + OD > BD ... (ii). Adding both inequalities: OA + OC + OB + OD > AC + BD. Therefore, OA + OB + OC + OD > AC + BD.
Answer: To establish this inequality, we apply the triangle inequality property. In triangle ACO, the sum of any two sides is greater than the third side. Therefore OA + OC > AC. Similarly, in triangle BDO, OB + OD > BD. Adding these two inequalities together gives us OA + OC + OB + OD > AC + BD, which rearranges to OA + OB + OC + OD > AC + BD.
In simple words: The path from O to all four corners is always longer than the two diagonals. This follows from the basic triangle rule that any two sides of a triangle are longer than the third side.

Exam Tip: Break a complex geometric inequality into simpler triangle inequalities - this simplification often makes the proof clear.

Question 9. Given: ABCD is a quadrilateral and AC is one of its diagonals. Given: A square ABCD in which ∠PQR = 90° and PB = QC = DR. To Prove: (i) QB = RC, (ii) PQ = QR, (iii) ∠QPR = 45°
Answer: (i) Consider the line segment QB. We have QB = BC - QC = CD - DR (since ABCD is a square, BC = DC and QC = DR are given). Thus QB = RC.

(ii) In △PBQ and △QCR, we have PB = QC (given), ∠PBQ = ∠QCR = 90° (all angles in a square), and QB = RC (from part i). By Side-Angle-Side criterion, △PBQ ≅ △QCR. Therefore, PQ = QR.

(iii) Since PQ = QR and ∠PQR = 90°, triangle PQR is an isosceles right triangle. The base angles are equal, and their sum is 90°. Thus, ∠QPR = ∠QRP = 90°/2 = 45°.
In simple words: Equal segments on the sides of a square create congruent triangles. An isosceles right triangle always has 45° base angles.

Exam Tip: In square configurations with equal segments, check for congruent right triangles that may be isosceles.

Question 10. Given: ABCD is a quadrilateral. To Prove: (i) AB + BC + CD + DA > 2AC, (ii) AB + BC + CD > DA, (iii) AB + BC + CD + DA > AC + BD
Answer: (i) In △ABC, by the triangle inequality, AB + BC > AC. In △ACD, by the triangle inequality, AD + CD > AC. Adding both inequalities: AB + BC + CD + DA > 2AC.

(ii) In △ABC, AB + BC > AC. On adding CD to both sides: AB + BC + CD > AC + CD. Since AC + CD > DA (triangle inequality in △ACD), we have AB + BC + CD > DA.

(iii) From (i), AB + BC + CD + DA > 2AC can be rewritten. Similarly, considering diagonal BD: AB + AD > BD and CB + CD > BD. Adding: AB + BC + CD + DA > 2BD. Combining both diagonal inequalities yields AB + BC + CD + DA > AC + BD.
In simple words: Each side of a quadrilateral forms triangles with the diagonals. The triangle inequality applied to these triangles proves all three statements.

Exam Tip: Always apply the triangle inequality - it is the foundation for quadrilateral inequalities.

Exercise 9B

Question 1. The figure shows quadrilateral ABCD with diagonals AC and BD intersecting. 
Answer: To prove the three inequalities: (i) AB + BC + CD + DA > 2AC. In △ABC, AB + BC > AC, and in △ACD, AD + CD > AC. Adding these gives the result. (ii) AB + BC + CD > DA. From △ABC, AB + BC > AC. Adding CD and using AC + CD > DA from △ACD yields the result. (iii) AB + BC + CD + DA > AC + BD. Applying triangle inequality to both diagonals and combining gives this result.
In simple words: Use the fact that in any triangle, the sum of two sides exceeds the third. Apply this to triangles formed by the diagonals.

Exam Tip: Break the quadrilateral into triangles using the diagonals, then apply triangle properties systematically.

Question 2. In a parallelogram ABCD, ∠A = 72°. Find ∠B, ∠C, and ∠D.
Answer: In a parallelogram, opposite angles are equal. Thus ∠A = ∠C = 72°. The sum of all four angles equals 360°. So ∠A + ∠B + ∠C + ∠D = 360°. Since ∠B = ∠D (opposite angles), we have 72° + ∠B + 72° + ∠B = 360°. This gives 144° + 2∠B = 360°, so 2∠B = 216°, and ∠B = 108°. Therefore ∠C = 72° and ∠D = 108°.
In simple words: In a parallelogram, opposite angles match, and all four angles add to 360°. Use these facts to find the unknown angles.

Exam Tip: In a parallelogram, remember that opposite angles are equal and adjacent angles are supplementary (sum to 180°).

Question 3. ABCD is a parallelogram with ∠A = 80° and ∠B = 60°. Find ∠ABD and ∠CDB.
Answer: ABCD is a parallelogram, so opposite angles are equal. Thus ∠C = ∠A = 80°. Since AD ∥ BC and BD is a transversal, the alternate angles are equal. Therefore ∠ADB = ∠DBC = 60°. In triangle ABD, ∠A + ∠ADB + ∠ABD = 180°. So 80° + 60° + ∠ABD = 180°, which gives ∠ABD = 40°. In a parallelogram, opposite angles are equal. So ∠ADC = ∠ABC = 100°. Therefore ∠CDB = ∠ADC - ∠ADB = 100° - 60° = 40°. Also, ∠ADB = 60° and ∠ADB = ∠CDB = 60° (alternate angles). Wait - let me recalculate: Since AB ∥ CD and AC is a transversal, ∠ACB = ∠DAC = 40° (alternate angles). So ∠ABC = ∠ABD + ∠DBC. We have 100° = ∠ABD + 60°, so ∠ABD = 40°. And in triangle BCD: ∠DBC + ∠BCD + ∠CDB = 180°, so 60° + 80° + ∠CDB = 180°, giving ∠CDB = 40°.
In simple words: Use parallel line properties to find alternate angles, then apply triangle angle sum to find the remaining angles.

Exam Tip: When diagonals divide a parallelogram, always use the parallel line properties and triangle angle sums together.

Question 4. ABCD is a parallelogram in which DA = 60° and bisectors of ∠A and ∠B meet at P. To Prove: (i) ∠APB = 90°, (ii) DP = PC, (iii) DC = 2AD.
Answer: (i) Since AB ∥ DP and AP is a transversal, ∠APD = ∠PAB = 30° (alternate angles). Also, AB ∥ PC and BP is a transversal, so ∠ABP = ∠CPB = 60° (alternate angles). But ∠ABP = ∠CPB = 60°. In triangle APB, ∠PAB + ∠ABP + ∠APB = 180°. So 30° + 60° + ∠APB = 180°, giving ∠APB = 90°.

(ii) Since ∠APD = ∠DAP = 30°, triangle APD is isosceles with DP = AD. Since ∠CPB = ∠CBP = 60°, triangle BPC is isosceles with PC = BC. Since ABCD is a parallelogram, BC = AD. Therefore DP = AD = BC = PC.

(iii) Since ∠APD = ∠PAB and ∠DAP = ∠APD = 30°, and similarly ∠CPB = ∠CBP = 60°. From the construction shown, P lies on DC. Since DP = PC, P is the midpoint of DC. Therefore DC = 2AD.
In simple words: Angle bisectors in a parallelogram create special triangles. Equal angles in these triangles make them isosceles, which gives the required relationships.

Exam Tip: When angle bisectors are involved, find the angles they create and check for isosceles triangles.

Question 5. ABCD is a parallelogram. Given: ∠AOB = 35°, ∠COD = 105°. To Find: (i) ∠ABO, (ii) ∠ODC, (iii) ∠CBD
Answer: (i) ∠AOB = ∠COD = 105° (vertically opposite angles). Now in triangle AOB, ∠OAB + ∠AOB + ∠ABO = 180°. So 35° + 105° + ∠ABO = 180°, giving ∠ABO = 40°.

(ii) Since AB ∥ DC and BD is a transversal, ∠ABD = ∠CDB (alternate angles). So ∠CDB = ∠ABD = 40°. Therefore ∠ODC = 40°.

(iii) As AB ∥ CD and AC is a transversal, ∠ACB = ∠DAC = 40° (alternate angles). Also, ∠CAB = ∠ACD = 40° (alternate angles). In triangle ABC, ∠CAB + ∠ABC + ∠ACB = 180°. From the parallelogram, ∠ABC = 180° - ∠BAD = 180° - (35° + 40°) = 105°. Wait - recalculating: In the diagram, we see ∠AOB = 35° at point O inside. From the isosceles triangle properties and using ∠ABO = 40°, ∠ABD = 40°, and ∠DBC = ∠ABC - ∠ABD = 65°. So ∠CBD = 65°.
In simple words: Use vertically opposite angles and the angle sum in triangles. Apply parallel line properties to transfer angles between triangles.

Exam Tip: Track which angles are vertically opposite and which are alternate - this keeps the solution organized.

Question 6. In a parallelogram ABCD, suppose ∠A = x°. Then ∠B, which is the adjacent angle, is (4/5)x°. Find all angles of the parallelogram.
Answer: In a parallelogram, adjacent angles are supplementary (sum to 180°). So ∠A + ∠B = 180°. Substituting ∠A = x° and ∠B = (4/5)x°, we get x + (4/5)x = 180°. This simplifies to (9/5)x = 180°, so x = 100°. Therefore ∠A = 100°, ∠B = (4/5) × 100 = 80°. In a parallelogram, opposite angles are equal, so ∠C = ∠A = 100° and ∠D = ∠B = 80°.
In simple words: Adjacent angles in a parallelogram always add up to 180°. Set up an equation and solve for x, then use the angle value to find all four angles.

Exam Tip: Always check that your four angles sum to 360° and that opposite angles are equal.

Question 8. Perimeter of a parallelogram ABCD = AB + BC + CD + DA = 9.5 + BC + 9.5 + BC (since ABCD is a parallelogram and its opposite sides are equal, i.e., AB = CD and BC = DA). Given: Perimeter = 30 cm. So 9.5 + BC + 9.5 + BC = 30. This gives 19 + 2BC = 30, so 2BC = 11, and BC = 11/2 = 5.5 cm. Therefore AB = 9.5 cm, BC = 5.5 cm, CD = 9.5 cm, DA = 5.5 cm.
Answer: The perimeter is the sum of all four sides. In a parallelogram, opposite sides are equal, so AB = CD and BC = DA. The perimeter equals AB + BC + CD + DA = 9.5 + BC + 9.5 + BC. Given that the perimeter is 30 cm, we have 19 + 2BC = 30, which gives 2BC = 11, so BC = 5.5 cm. Therefore the sides are AB = 9.5 cm, BC = 5.5 cm, CD = 9.5 cm, and DA = 5.5 cm.
In simple words: In a parallelogram, opposite sides are the same length. Add them up using the perimeter formula, then solve for the unknown side.

Exam Tip: Always use the property that opposite sides of a parallelogram are equal when finding side lengths.

Question 9. (i) ABCD is a rhombus, so all its sides are equal. In triangle ABC, we have AB = BC, so ∠CAB = ∠ACB = x°. As, ∠CAB + ∠ABC + ∠ACB = 180°, we get x + 110° + x = 180°, which gives 2x = 70°, so x = 35°. And y = 35°
Answer: Since ABCD is a rhombus, all four sides are equal in length. In triangle ABC, AB = BC, making it isosceles. Therefore, the base angles are equal: ∠CAB = ∠ACB = x°. The sum of angles in the triangle is 180°, so x + 110° + x = 180°, which gives 2x = 70°. Thus x = 35°. Similarly, y = 35°.
In simple words: A rhombus has all equal sides. Equal sides in a triangle make equal opposite angles. Use angle sum to find them.

Exam Tip: In a rhombus, recognize isosceles triangles formed by the diagonals and sides.

Question 10. (i) Since in a rhombus, all sides are equal. So in triangle ABD, AB = AD, and ∠ABD = ∠ADB. Now, x = y ... (1). Now in triangle ABC, AB = BC, so ∠CAB = ∠ACB. So ∠ACB = 40°. Now, ∠B = 180° - ∠CAB - ∠ACB = 180° - 40° - 40° = 100°. So, ∠DBC = ∠B - x° = 100° - x°. But ∠DBC = ∠ADB = y° (alternate angle). So, 100° - x° = y°. From (1), 100° - x° = x°, which gives 100° = 2x°, so x° = 50° and y° = 50°. (ii) Since ABCD is a rhombus. So, ∠A = ∠C, i.e., ∠C = 62°. Now in triangle BCD, BC = DC, so ∠CDB = ∠DBC = y°. As, ∠BDC + ∠DBC + ∠BCD = 180°. So, y + y + 62° = 180°, giving 2y = 180° - 62° = 118°, so y = 59°. As diagonals of a rhombus are perpendicular to each other, triangle COD is a right triangle and ∠DOC = 90°, ∠ODC = y = 59°. So, ∠DCO = 90° - ∠ODC = 90° - 59° = 31°. Therefore, x = 31° and y = 59°.
Answer: (ii) In a rhombus, opposite angles are equal. So ∠A = ∠C = 62°. In triangle BCD, since BC = DC (all sides of a rhombus are equal), the triangle is isosceles. The base angles are equal: ∠CDB = ∠DBC = y°. Using the angle sum property, y + y + 62° = 180°, which gives 2y = 118°, so y = 59°. The diagonals of a rhombus intersect at right angles. In the right triangle COD, ∠DOC = 90° and ∠ODC = 59°. Therefore ∠DCO = 90° - 59° = 31°. Thus x = 31° and y = 59°.
In simple words: Rhombuses have equal opposite angles and perpendicular diagonals. These properties, combined with isosceles triangles, determine all the angles.

Exam Tip: In rhombus problems, always use the perpendicular diagonals property and the equal opposite angles property.

Question 11. Since diagonals of a rhombus bisect each other at right angles. So AO = OC = (1/2)AC = (1/2) × 16 = 8 cm. In right triangle AOB, AB² = AO² + OB². So 10² = 8² + OB², which gives OB² = 100 - 64 = 36, so OB = 6 cm. Length of the other diagonal BD = 2 × OB = 2 × 6 = 12 cm.
Answer: The diagonals of a rhombus bisect each other at right angles. So AO = (1/2)AC = (1/2) × 16 = 8 cm. In the right triangle AOB formed by half of each diagonal and a side, we have AB² = AO² + OB². Substituting the known values: 10² = 8² + OB², so 100 = 64 + OB², giving OB² = 36 and OB = 6 cm. The full length of diagonal BD is BD = 2 × OB = 2 × 6 = 12 cm.
In simple words: Rhombus diagonals split each other at 90°. Use the Pythagorean theorem in the right triangle formed to find the other diagonal.

Exam Tip: Always apply the right angle property when using the Pythagorean theorem in rhombus diagonal problems.

Question 12. ABCD is a rhombus in which diagonal AC = 24 cm and BD = 18 cm. We know that in a rhombus, diagonals bisect each other at right angles. So in triangle AOB, ∠AOB = 90°, AO = (1/2)AC = (1/2) × 24 = 12 cm, and BO = (1/2)BD = (1/2) × 18 = 9 cm. By Pythagoras Theorem, we have AB² = AO² + OB². So AB² = 12² + 9² = 144 + 81 = 225, and AB = √225 = 15 cm. So the length of each side of the rhombus is 15 cm. Area of triangle ABC = (1/2) × AC × OB = (1/2) × 16 × 6 = 48 cm². Area of triangle ACD = (1/2) × AC × OD = (1/2) × 16 × 6 = 48 cm². Therefore, Area of rhombus ABCD = (Area of △ABC + Area of △ACD) = (48 + 48) cm² = 96 cm².
Answer: The diagonals of a rhombus bisect each other at right angles. In triangle AOB, AO = (1/2) × 24 = 12 cm and BO = (1/2) × 18 = 9 cm, with ∠AOB = 90°. By the Pythagorean theorem, AB² = 12² + 9² = 144 + 81 = 225, so AB = 15 cm. The area of triangle ABC equals (1/2) × AC × BO = (1/2) × 24 × 9 = 108 cm². Similarly, the area of triangle ACD = (1/2) × AC × DO = (1/2) × 24 × 9 = 108 cm². The total area of the rhombus is 108 + 108 = 216 cm². Note: using the formula Area = (1/2) × d₁ × d₂ = (1/2) × 24 × 18 = 216 cm².
In simple words: The diagonals split the rhombus into four right triangles. Find the side length using the Pythagorean theorem, then compute area using the diagonal formula.

Exam Tip: For a rhombus, always use the formula Area = (1/2) × d₁ × d₂ where d₁ and d₂ are the diagonals.

Question 13. (i) We know that diagonals of a rectangle are equal and bisect each other. So, in triangle AOB, AO = OB. So, ∠OAB = ∠OBA (base angles are equal). i.e., ∠OBA = 35°. ∠AOB = 180° - 35° - 35° = 110°. And, ∠DOC = y° = ∠AOB = 110° (Vertically opposite angles). Consider the right triangle, triangle ABC, right angled at B. So, ∠ABC = 90°. Now, consider the triangle OBC. So, ∠OBC = x° = ∠ABC - ∠OBA = 90° - 35° = 55°. Therefore x = 55° and y = 110°. (ii) We know that diagonals of a rectangle are equal and bisect each other. So, in triangle AOB, OA = OB, so ∠OAB = ∠OBA. Again in triangle AOB, ∠AOB + ∠OAB + ∠OBA = 180°. So, 110° + ∠OAB + ∠OBA = 180°. So, 2∠OAB = 180° - 110° = 70°. So, ∠OAB + ∠OBA = 35°. Since AB || CD and AC is a transversal, ∠DCA and ∠CAB are alternate angles, and thus they are equal. So, ∠DCA = y° = ∠CAB and ∠CAB = 35° .... (1). Now consider the right triangle, triangle ABC. ∠ACB = x° = 90° - ∠CAB = 90° - 35° = 55°. Therefore x = 55° and y = 35°.
Answer: (i) In a rectangle, diagonals are equal and bisect each other. In triangle AOB, since AO = OB, the triangle is isosceles, so ∠OAB = ∠OBA = 35°. Thus ∠AOB = 180° - 35° - 35° = 110°. Vertically opposite angles are equal, so ∠DOC = 110°. In right triangle ABC (∠ABC = 90°), we find ∠OBC = ∠ABC - ∠OBA = 90° - 35° = 55°. Therefore x = 55° and y = 110°.

(ii) In triangle AOB (isosceles since OA = OB), if ∠AOB = 110°, then ∠OAB + ∠OBA = 70°. Since these base angles are equal, ∠OAB = ∠OBA = 35°. Since AB ∥ CD and AC is a transversal, ∠DCA = ∠CAB = 35° (alternate angles). In right triangle ABC, ∠ACB = 90° - 35° = 55°. Therefore x = 55° and y = 35°.
In simple words: Rectangle diagonals are equal and split each other in half. Use isosceles triangle properties and parallel line properties to find angles.

Exam Tip: In rectangle diagonal problems, always recognize the isosceles triangles formed and use alternate angle properties.

Question 14. A parallelogram ABCD in which AL and CM are perpendiculars to its diagonal BD. To Prove: (i) △ALD ≅ △CMB, (ii) AL = CM
Answer: (i) In triangles ALD and CMB, we have ∠ALD = ∠CMB = 90° (given that AL ⊥ BD and CM ⊥ BD). Since AD ∥ BC and BD is a transversal, ∠ADL = ∠CBM (alternate angles). The side AD = BC (opposite sides of a parallelogram are equal). By Angle-Angle-Side criterion, △ALD ≅ △CMB.

(ii) Since the triangles are congruent, their corresponding parts are equal. Therefore, AL = CM.
In simple words: Two triangles formed by perpendiculars from opposite vertices to a diagonal in a parallelogram are congruent. This proves the perpendicular segments are equal.

Exam Tip: In parallelogram problems with perpendiculars, use alternate angles and the equal opposite sides property to establish congruence.

Question 16. A parallelogram ABCD in which the angle bisectors of ∠A and ∠B intersect at P. Prove that ∠APB = 90°
Answer: Since AD and BC are parallel and AB serves as a transversal, the sum of co-interior angles is 180°. Therefore, ∠A + ∠B = 180°. The angle bisectors divide these angles in half: ∠PAB = (1/2)∠A and ∠PBA = (1/2)∠B. In triangle PAB, the sum of all three angles equals 180°:

\[ \angle PAB + \angle PBA + \angle APB = 180° \]

\[ \frac{1}{2}\angle A + \frac{1}{2}\angle B + \angle APB = 180° \]

\[ \frac{1}{2}(\angle A + \angle B) + \angle APB = 180° \]

\[ \frac{1}{2}(180°) + \angle APB = 180° \]

\[ 90° + \angle APB = 180° \]

\[ \angle APB = 90° \]

Exam Tip: Always use the property that co-interior angles on a transversal cutting parallel lines sum to 180°. Recognise when angle bisectors are involved and apply the half-angle substitution in the triangle angle sum equation.

Question 17. A parallelogram ABCD in which AP = (1/3)AD and CQ = (1/3)BC. Prove that PAQC is a parallelogram.
Answer: In triangles ABQ and CDP, we observe that AB = CD (opposite sides of a parallelogram). ∠B = ∠D (opposite angles of a parallelogram). We also compute:

DP = AD - PA = AD - (1/3)AD = (2/3)AD

BQ = BC - CQ = BC - (1/3)BC = (2/3)BC

Since AD = BC (opposite sides of a parallelogram), we have DP = BQ = (2/3)AD. By the Side-Angle-Side congruence criterion, △ABQ ≅ △CDP. Consequently, AQ = CP. Since ∠CAE = ∠DCA (alternate interior angles where line AC is a transversal) and ∠QAP = ∠A - ∠QAB and ∠QCP = ∠C - ∠PCD, and these differences are equal (because ∠A = ∠C in a parallelogram and from equation (1) the relevant angle measures match), we conclude that AQ ∥ CP. With PA = (1/3)AD and CQ = (1/3)BC = (1/3)AD, we have PA = CQ. Thus, PAQC has AQ = CP and PA = CQ with these opposite sides parallel and equal, making PAQC a parallelogram.

Exam Tip: When given fractional divisions of sides in a parallelogram, calculate the remaining segments and use congruence of triangles to establish equality of opposite sides. Verify parallelism using alternate interior angles or direct comparison of angle measures.

Question 18. A parallelogram ABCD in which the diagonals intersect at O. E and F are points on AB and CD respectively. Prove that OE = OF.
Answer: In triangles AOE and COF, we have:

∠CAE = ∠DCA (alternate angles, since AC acts as a transversal across the parallel lines AB and CD)

AO = CO (the diagonals of a parallelogram bisect each other)

∠AOE = ∠COF (vertically opposite angles)

By the Angle-Side-Angle congruence criterion, △AOE ≅ △COF. The matching parts of the congruent triangles are equal, which gives us OE = OF.

Exam Tip: Always use the key property that diagonals of a parallelogram bisect each other. Combine this with alternate interior angles and vertically opposite angles to establish triangle congruence by ASA.

Question 19. ABCD is a parallelogram in which AB is extended to E such that BE = AB. DE is joined and cuts BC at O. Prove that OB = OC.
Answer: In triangles OCD and OBE, we observe:

∠DOC = ∠EOB (vertically opposite angles)

∠OCD = ∠OBE (AB and CD are parallel, with BC acting as a transversal, so alternate angles are equal)

DC = BE (since DC = AB and BE = AB, we have DC = BE)

By the Angle-Angle-Side congruence criterion, △OCD ≅ △OBE. Consequently, the corresponding sides of the congruent triangles are equal, yielding OC = OB. Therefore, OB = OC.

Exam Tip: Identify vertically opposite angles at the intersection point and use the parallel line property to find alternate angles. Notice when equal segments combine with the parallelogram's properties to establish congruence.

Question 20. A parallelogram ABCD in which E is the midpoint of side BC. DE and AB when extended meet at F. Prove that AF = 2AB.
Answer: In triangles DEC and FEB, we note:

∠DEC = ∠FEB (vertically opposite angles)

∠DCE = ∠FBE (alternate angles, since CD ∥ AB with EB serving as a transversal)

CE = EB (E is the midpoint of BC)

By the Angle-Angle-Side congruence criterion, △DEC ≅ △FEB. The corresponding sides are equal, so DC = FB. Since ABCD is a parallelogram, DC = AB. Therefore, FB = AB. Now, AF = AB + BF = AB + DC = AB + AB = 2AB. Thus, AF = 2AB.

Exam Tip: When a midpoint divides a side, use it to establish equal segments that appear in triangle congruence. Apply the parallelogram property (opposite sides are equal) to connect the congruent segment to the original sides.

Question 21. A triangle ABC in which through points A, B and C, lines QR, QP and RP are drawn parallel to BC, CA and AB respectively.
Answer: Since AR ∥ BC and AB ∥ RC (given), ABCR is a parallelogram. Thus, AR = BC. Also, since AQ ∥ BC and QB ∥ AC (given), AQBC is a parallelogram. Thus, QA = BC. Adding the two results, AR + QA = BC + BC, which gives QR = 2BC. Therefore, BC = (1/2)QR. Similarly, through analogous arguments for the other pairs of parallel lines, we can show AB = (1/2)RP and AC = (1/2)PQ. The triangle PQR has all its sides equal to twice the corresponding sides of triangle ABC, confirming the construction and the perimeter relationship: Perimeter of △PQR = 2(Perimeter of △ABC).

Exam Tip: Recognise when parallel lines through vertices create parallelograms with the original triangle. Use opposite sides of these parallelograms to relate the new triangle's sides to the original triangle's sides.

Exercise 9C

Question 1. A triangle ABC in which through points A, B and C, lines QR, QP and RP are drawn parallel to BC, CA and AB respectively. Prove that BC = (1/2)QR.
Answer: Since AR ∥ BC and AB ∥ RC (by construction), ABCR forms a parallelogram. In a parallelogram, opposite sides are equal, so AR = BC. Similarly, since AQ ∥ BC and QB ∥ AC (by construction), AQBC forms a parallelogram. Thus, QA = BC. Adding equations (i) and (ii): AR + QA = BC + BC, which simplifies to QR = 2BC. Therefore, dividing both sides by 2 gives us BC = (1/2)QR.

Exam Tip: Draw all three parallel lines carefully to visualise the parallelograms formed. Use the property that opposite sides of a parallelogram are equal to establish the required relationships.

Question 2. A trapezium ABCD in which AB ∥ DC and through the mid-point E of AD a line is drawn parallel to AB which cuts BC at F. Prove that F is the mid-point of BC.
Answer: Since AB ∥ DC and EF ∥ AB, it follows that AB ∥ EF ∥ DC. By the Intercept Theorem, if three parallel lines intersect a transversal and produce equal intercepts on one transversal, then the intercepts on any other transversal are also equal. Applying this theorem to transversal AD: the intercepts made by lines AB, EF and DC are equal. Specifically, AE = ED (since E is the midpoint of AD). Now, AD serves as a transversal cutting the three parallel lines. Therefore, by the Intercept Theorem applied to transversal BC, the corresponding intercepts are CF = FB, showing that F is the midpoint of BC.

Exam Tip: The Intercept Theorem is the key tool here. Understand that equal spacing on one transversal guarantees equal spacing on any parallel transversal cutting the same set of parallel lines.

Question 3. A parallelogram ABCD in which E and F are the mid-points of AB and CD. A line segment GH cuts EF at P.
Answer: In this configuration, AD, EF and BC are three parallel line segments (since E and F are midpoints of opposite sides of the parallelogram, EF ∥ AD and EF ∥ BC). The intercepts made by the three lines on transversal AB are equal because AE = EB (E is the midpoint). By the Intercept Theorem, when GH intersects these three parallel segments, the intercepts on GH are also equal. Since DF = FC (F is the midpoint of CD), and by the Intercept Theorem applied to transversal GH cutting the parallel lines AD, EF and BC, we have GP = PH. Therefore, the line GH is bisected by EF at point P.

Exam Tip: Midpoints of opposite sides in a parallelogram create a line segment parallel to the other two sides. Use the Intercept Theorem to show equal intercepts on any transversal.

Question 4. A triangle ABC in which AD is its median and DE ∥ AB.
Answer: In triangle ABC, the line drawn through the midpoint of one side parallel to another side intersects the third side at its midpoint. Since D is the midpoint of BC (as AD is a median) and DE ∥ AB, the line DE intersects AC at its midpoint E. By the Midpoint Theorem, E is the midpoint of AC. Therefore, BE is the median of triangle ABC drawn through vertex B. This establishes that BE is a median of triangle ABC.

Exam Tip: Remember the Midpoint Theorem: a line segment connecting the midpoints of two sides of a triangle is parallel to the third side and equal to half its length. Use this to identify other medians or midpoints.

Question 5. A trapezium ABCD in which AB ∥ DC. P and Q are the mid-points of AD and BC. DQ is joined and produced and AB is also produced and so that they meet at E. AC cuts PQ at R.
Answer: (i) In triangles QCD and QBE, we have ∠DQC = ∠EQB (vertically opposite angles). Since Q is the midpoint of BC, we have CQ = BQ. ∠QCD = ∠QEB (alternate angles, as DC ∥ AB). By Angle-Side-Angle congruence, △QCD ≅ △QBE. The corresponding parts of congruent triangles are equal, so DQ = QE.

(ii) By the Midpoint Theorem, since PQ connects the midpoints P and Q of sides AD and BC respectively, and DC and AB are parallel, PQ ∥ AB and PQ ∥ DC. Since the intercepts made by AB, PQ and DC on transversal AD are equal (AP = PD), by the Intercept Theorem on transversal BC, the intercepts are also equal. Thus, PR ∥ AB ∥ DC.

(iii) By the Intercept Theorem, if three parallel lines cut a transversal, equal intercepts on one transversal imply equal intercepts on any other transversal. Since AR = RC (established through the parallel line properties and intercept equality), AC is bisected at R by the line PQ.

Exam Tip: Use triangle congruence to establish equal segments. Apply the Intercept Theorem systematically to relate intercepts on different transversals cutting parallel lines.

Question 6. A parallelogram ABCD in which E is the mid-point of DC. A line is drawn through D parallel to EB meeting AB at F and BC produced at G.
Answer: (i) In triangle CDG, EB ∥ DG (by construction) and E is the midpoint of CD. By the Midpoint Theorem, the line segment connecting a point that divides one side and parallel to another divides the third side proportionally. Applying this, B is the midpoint of CG. Since ABCD is a parallelogram, AD = BC. Given that E is the midpoint of DC, we have AD = BG = (1/2)CG. Therefore, AD = (1/2)GC.

(ii) In triangle CDG, since E is the midpoint of DC and B is the midpoint of CG (from part i), by the Midpoint Theorem applied in reverse, DG = 2EB. Therefore, DG = 2EB.

Exam Tip: The Midpoint Theorem applies in multiple forms - use the standard form when you know a midpoint and a parallel line, and use the converse when you need to find relationships between segments.

Question 7. A triangle ABC in which D, E and F are the mid-points of BC, AC and AB respectively. DE, EF and FD are joined to get four triangles.
Answer: Since F, E are midpoints of AB and AC, by the Midpoint Theorem, EF ∥ BC and EF = (1/2)BC. Similarly, FD ∥ AC and FD = (1/2)AC. Also, ED ∥ AB and ED = (1/2)AB. In triangles AFE and BFD, we have AF = FB (F is the midpoint of AB), FE = (1/2)BC = BD (since D is the midpoint of BC, BD = (1/2)BC, and FE = (1/2)BC), and FD = (1/2)AC = AE (since E is the midpoint of AC, AE = (1/2)AC, and FD = (1/2)AC). By Side-Side-Side congruence, △AFE ≅ △BFD. Similarly, we can show △AFE ≅ △FDE and △BFD ≅ △EDC. Therefore, all four triangles AFE, BFD, FDE and EDC are congruent to each other.

Exam Tip: When midpoints divide all three sides of a triangle, the resulting four triangles are always congruent. This follows directly from the Midpoint Theorem and the congruence of the segments created.

Question 8. A triangle ABC in which D, E and F are the mid-points of BC, AC and AB respectively. DE, EF and FD are joined to get four triangles.
Answer: Since D, E, F are the midpoints of BC, AC and AB respectively, by the Midpoint Theorem, the segments DE, EF and FD are parallel to the sides and equal to half of them:

EF ∥ BC and EF = (1/2)BC
FD ∥ AC and FD = (1/2)AC
ED ∥ AB and ED = (1/2)AB

In triangles AFE and BFD, we have:
AF = FB = (1/2)AB (F is the midpoint)
FE = (1/2)BC (from Midpoint Theorem)
AE = EC = (1/2)AC (E is the midpoint)

Also, BD = DC = (1/2)BC (D is the midpoint)

Since FE = BD = (1/2)BC and FD = (1/2)AC = AE, and AF = FB, by the Side-Side-Side criterion, △AFE ≅ △BFD. By similar reasoning applied to the other pairs, all four triangles are congruent to each other.

Exam Tip: Always apply the Midpoint Theorem when midpoints are identified. The resulting smaller triangle formed by joining the three midpoints divides the original triangle into four congruent triangles.

Question 9. A triangle ABC in which D, E and F are the mid-points of BC, AC and AB respectively.
Answer: We need to prove that ∠EDF = ∠A, ∠DEF = ∠B, and ∠DFE = ∠C. By the Midpoint Theorem, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half its length. In triangle ABC, EF ∥ BC and EF = (1/2)BC. FD ∥ AC and FD = (1/2)AC. ED ∥ AB and ED = (1/2)AB. Since EF ∥ BC, the angles formed by EF with other segments match the angles formed by BC with those segments. Specifically, ∠FED = ∠ABC (corresponding angles or alternate angles depending on the transversal). Similarly, ∠EDF = ∠CAB (since FD ∥ AC and using the properties of parallel lines cut by a transversal ED). And ∠EFD = ∠ACB (since ED ∥ AB and using the properties of parallel lines cut by a transversal FD). Therefore, triangle DEF is similar to triangle ABC with ∠EDF = ∠A, ∠DEF = ∠B, and ∠DFE = ∠C.

Exam Tip: When the Midpoint Theorem creates parallel segments, use the properties of parallel lines cut by transversals to relate angles in the new configuration to angles in the original triangle.

Question 10. A rectangle ABCD in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.
Answer: We join the diagonals AC and BD. In triangle ABC, P and Q are the midpoints of AB and BC respectively. By the Midpoint Theorem, PQ ∥ AC and PQ = (1/2)AC. Similarly, in triangle BCD, Q and R are midpoints, so QR ∥ BD and QR = (1/2)BD. In triangle ACD, R and S are midpoints, so RS ∥ AC and RS = (1/2)AC. In triangle ABD, S and P are midpoints, so SP ∥ BD and SP = (1/2)BD. The diagonals of a rectangle are equal: AC = BD. Therefore, PQ = RS = (1/2)AC and QR = SP = (1/2)BD. Since AC = BD, all four sides PQ, QR, RS and SP are equal. Additionally, since PQ ∥ RS (both parallel to AC) and QR ∥ SP (both parallel to BD), PQRS is a parallelogram with all sides equal. In a rectangle, the diagonals are perpendicular at the midpoint and intersect at right angles, so ∠EOF = 90°. Therefore, ∠QOR = 90°, making PQRS a rhombus.

Exam Tip: Use the Midpoint Theorem on each of the four triangles formed by the rectangle's diagonals. Equal diagonals in a rectangle ensure that the quadrilateral formed by joining the midpoints is a rhombus with all sides equal.

Question 11. A rhombus ABCD in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.
Answer: We join the diagonals AC and BD. In triangle ABC, P and Q are midpoints of AB and BC. By the Midpoint Theorem, PQ ∥ AC and PQ = (1/2)AC. Similarly, in triangle BCD, Q and R are midpoints, so QR ∥ BD and QR = (1/2)BD. In triangle ACD, R and S are midpoints, so RS ∥ AC and RS = (1/2)AC. In triangle ABD, S and P are midpoints, so SP ∥ BD and SP = (1/2)BD. Therefore, PQ = RS = (1/2)AC and QR = SP = (1/2)BD. In a rhombus, the diagonals are equal in length. We have AC = BD, so PQ = QR = RS = SP = (1/2)AC. Since all four sides are equal and opposite sides are parallel (PQ ∥ RS, QR ∥ SP), PQRS is a parallelogram with all sides equal. Since PQ ∥ AC and QR ∥ BD, and the diagonals of a rhombus are perpendicular, ∠PQR = 90°. Therefore, PQRS is a rectangle.

Exam Tip: In a rhombus, the diagonals are perpendicular. When you apply the Midpoint Theorem to form a quadrilateral from the midpoints, the perpendicularity of the diagonals translates to perpendicular adjacent sides in PQRS, making it a rectangle.

Question 12. A square ABCD in which E, F, G and H are the mid-points of AB, BC, CD and AD respectively. The mid-points are joined together.
Answer: We join diagonals AC and BD. In triangle ABC, E and F are midpoints of AB and BC. By the Midpoint Theorem, EF ∥ AC and EF = (1/2)AC. Similarly, in triangle BCD, F and G are midpoints, so FG ∥ BD and FG = (1/2)BD. In triangle ACD, G and H are midpoints, so GH ∥ AC and GH = (1/2)AC. In triangle ABD, H and E are midpoints, so HE ∥ BD and HE = (1/2)BD. Since the diagonals of a square are equal (AC = BD) and perpendicular, we have EF = FG = GH = HE = (1/2)AC. Also, EF ∥ GH (both parallel to AC) and FG ∥ HE (both parallel to BD). Thus, EFGH is a parallelogram with all sides equal, making it a rhombus. Since AC ⊥ BD (diagonals are perpendicular in a square), and EF is parallel to AC while FG is parallel to BD, we have EF ⊥ FG, so ∠EFG = 90°. In a parallelogram where one angle is 90°, all angles are 90°. Therefore, EFGH is a square.

Exam Tip: For a square, both diagonals are equal and perpendicular. The quadrilateral formed by joining the midpoints inherits both properties - equal sides from equal diagonals and right angles from perpendicular diagonals - making it also a square.

Question 13. A quadrilateral ABCD in which H, L, G and K are the mid-points of AB, BC, CD and AD respectively. Points G and H are joined and K and L are joined.
Answer: Construction: Join the diagonals KH, BD and GL. Proof: Since K and H are midpoints of AD and AB respectively, in triangle ABD, by the Midpoint Theorem, KH ∥ BD and KH = (1/2)BD. Similarly, since L and G are midpoints of BC and CD respectively, in triangle BCD, by the Midpoint Theorem, LG ∥ BD and LG = (1/2)BD. Therefore, KH ∥ LG and KH = LG. Now, in triangle ABC, H and L are midpoints of AB and BC, so by the Midpoint Theorem, HL ∥ AC and HL = (1/2)AC. Similarly, in triangle ACD, K and G are midpoints of AD and CD, so KG ∥ AC and KG = (1/2)AC. Therefore, HL ∥ KG and HL = KG. Since KHGL has both pairs of opposite sides parallel and equal (KH ∥ LG with KH = LG, and HL ∥ KG with HL = KG), KHGL is a parallelogram. Moreover, if the diagonals AC and BD of the original quadrilateral are equal (AC = BD), then HL = KG = KH = LG, and KHGL becomes a rhombus. If the diagonals are perpendicular, the angles of KHGL are right angles, making it a rectangle. If both conditions hold, KHGL is a square.

Exam Tip: The quadrilateral formed by joining the midpoints of any quadrilateral is always a parallelogram (Varignon's theorem). Its shape depends on the diagonals of the original quadrilateral - equal diagonals give a rhombus, perpendicular diagonals give a rectangle, and both give a square.