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🔷 Chapter 12: Quadrilaterals • Exercise Set 12.2

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Exercise Set 12.1 4Exercise Set 12.2 4Exercise Set 12.3 3Exercise Set 12.4 1End-of-Chapter Exercises 7All Exercises 19
Q1

1. True or false?

  1. (i) A parallelogram with a right angle is a rectangle.
  2. (ii) A rhombus with perpendicular diagonals is a square.
  3. (iii) If the diagonals of a parallelogram are equal, then it is a rectangle.
  • (i) True. If one angle is 90°, the opposite angle is also 90°. The adjacent angles must sum to 180°, making them 90° as well. A quadrilateral with four right angles is a rectangle.
  • (ii) False. Every rhombus has perpendicular diagonals. It only becomes a square if its diagonals are also equal in length (or if it has a right angle).
  • (iii) True. A parallelogram with equal diagonals guarantees that all four angles are 90°, satisfying the definition of a rectangle.
Q2

The diagonal AC of a parallelogram ABCD bisects ∠A. Show that it also bisects ∠C and that ABCD is a rhombus.

Proof:
1. Since ABCD is a parallelogram, AB || DC and AD || BC.
2. Given AC bisects ∠A, we have ∠DAC = ∠BAC.
3. Because AB || DC, the alternate interior angles are equal: ∠BAC = ∠DCA.
4. Similarly, because AD || BC, the alternate interior angles are equal: ∠DAC = ∠BCA.
5. By substitution, ∠DCA = ∠BCA. Thus, AC also bisects ∠C.
6. In ΔADC, since ∠DAC = ∠DCA, the sides opposite to equal angles are equal: AD = DC.
7. Since adjacent sides of the parallelogram are equal, all four sides must be equal (AB = DC = AD = BC). Therefore, ABCD is a rhombus.

A B C D
Diagonal AC bisects ∠A and ∠C ⇒ AD = DC ⇒ ABCD is a Rhombus
Q3

The following questions examine converses of true properties. Answer them with Yes or No. If your answer is No, what extra condition can you add so that the answer becomes Yes?

  • (i) If the diagonals of a quadrilateral ABCD bisect its angles, must ABCD be a rhombus?
  • (ii) If the diagonals of a quadrilateral ABCD bisect each other at right angles, must ABCD be a rhombus?
  • (iii) If the diagonals of a quadrilateral ABCD are of equal length, must ABCD be a rectangle?
  • (i) Yes. If diagonals bisect the angles, you can prove through triangle congruence (ASA) that adjacent sides are equal, making it a rhombus.
  • (ii) Yes. If the diagonals bisect each other, it is guaranteed to be a parallelogram. If they do so at right angles, it guarantees all four sides are equal, making it a rhombus.
  • (iii) No. An isosceles trapezium also has equal diagonals but is not a rectangle. Extra condition: Add "If the diagonals bisect each other and are of equal length..." then it becomes a rectangle.
Q4

Let ABCD be a parallelogram with AB ≠ BC. Show that the pairwise intersection points of the four angle bisectors form the vertices of a rectangle. Why did we assume AB ≠ BC?

Answer:
Let the bisectors of ∠A and ∠B intersect at P, ∠B and ∠C intersect at Q, ∠C and ∠D intersect at R, and ∠D and ∠A intersect at S.
1. In a parallelogram, consecutive angles sum to 180°. Therefore, ∠A + ∠D = 180°.
2. The bisectors of A and D form ΔASD. ∠DAS + ∠ADS = (∠A + ∠D)/2 = 180°/2 = 90°.
3. The third angle of the triangle, ∠ASD, must be 180° - 90° = 90°.
4. By the same logic, the intersections at P, Q, and R also form 90° angles. A quadrilateral PQRS with four right angles is a rectangle.
Why assume AB ≠ BC? If AB = BC, the parallelogram is a rhombus. In a rhombus, the angle bisectors are the diagonals themselves, meaning they intersect at exactly one point rather than forming a new four-sided region.

A B C D PQRS (90°)
Pairwise intersections of the four angle bisectors of a parallelogram form rectangle PQRS