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? (In the ancient Indian study of quadrilaterals, equality of diagonals was considered significant. Quadrilaterals were first classified according to whether their diagonals were equal or not, before considering equality of sides.)
Step-by-step solution
Idea: For each part, either prove the converse with congruent triangles, or find one shape (a counterexample) that satisfies the condition but is not the named shape.
(i) If the diagonals of a quadrilateral ABCD bisect its angles, must ABCD be a rhombus?
- AC bisects ∠A and ∠C. In ∆ABC and ∆ADC: ∠BAC = ∠DAC, AC common, ∠BCA = ∠DCA. So ∆ABC ≅ ∆ADC (ASA), giving AB = AD and CB = CD.1 mark
- BD bisects ∠B and ∠D. In the same way ∆BAD ≅ ∆BCD (ASA), giving BA = BC and DA = DC. So AB = BC = CD = DA: Yes, ABCD is a rhombus.1 mark
(ii) If the diagonals of a quadrilateral ABCD bisect each other at right angles, must ABCD be a rhombus?
- Diagonals bisect each other, so ABCD is a parallelogram. Let them meet at O.½ mark
- In ∆AOB and ∆AOD: OB = OD, ∠AOB = ∠AOD = 90°, AO common. So ∆AOB ≅ ∆AOD (SAS) and AB = AD. A parallelogram with adjacent sides equal has all sides equal: Yes, a rhombus.1 mark
(iii) If the diagonals of a quadrilateral ABCD are of equal length, must ABCD be a rectangle? (In the ancient Indian study of quadrilaterals, equality of diagonals was considered significant. Quadrilaterals were first classified according to whether their diagonals were equal or not, before considering equality of sides.)
- No. Counterexample: the isosceles trapezium A(−2, 3), B(2, 3), C(5, 0), D(−5, 0). AC = BD = √58, but ∠D ≈ 45°, so it is not a rectangle.1 mark
- Extra condition: the diagonals also bisect each other. Then ABCD is a parallelogram with equal diagonals, and such a parallelogram is a rectangle (∆ABC ≅ ∆BAD by SSS gives ∠A = ∠B = 90°).½ mark
Check: (iii) AC: from (−2, 3) to (5, 0): √(49 + 9) = √58; BD: from (2, 3) to (−5, 0): √(49 + 9) = √58 ✓. At D, side DA goes 3 right and 3 up, so ∠D = 45° ✓.
Answer to write in the exam
(i)
∆ABC ≅ ∆ADC (ASA: ∠BAC = ∠DAC, AC common, ∠BCA = ∠DCA) ⇒ AB = AD, CB = CD
∆BAD ≅ ∆BCD (ASA: ∠ABD = ∠CBD, BD common, ∠ADB = ∠CDB) ⇒ BA = BC, DA = DC
∴ AB = BC = CD = DA ⇒ rhombus. Yes.
(ii)
Diagonals bisect each other ⇒ ABCD is a parallelogram
∆AOB ≅ ∆AOD (SAS: OB = OD, ∠AOB = ∠AOD = 90°, AO common) ⇒ AB = AD
∴ Parallelogram with AB = AD ⇒ rhombus. Yes.
(iii)
Isosceles trapezium A(−2, 3), B(2, 3), C(5, 0), D(−5, 0): AC = BD = √58, ∠D = 45° ≠ 90° ⇒ not a rectangle
∴ No.
Extra condition: diagonals bisect each other ⇒ parallelogram with AC = BD ⇒ rectangle.
Common mistakes that cost marks
- In (i), using only one diagonal. A kite has one diagonal bisecting two angles but is not a rhombus; both diagonals are needed.
- In (ii), forgetting “bisect each other”: a kite has perpendicular diagonals but is not a rhombus.
- In (iii), answering yes from a picture of a rectangle. One counterexample (isosceles trapezium) settles it.
How this can come in the exam
Assertion (A): A quadrilateral whose diagonals are equal must be a rectangle.
Reason (R): The diagonals of a rectangle are equal.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Show answer
(D) A is false but R is true.
R is true, but A is false: an isosceles trapezium has equal diagonals and is not a rectangle. A true statement does not make its converse true.
Try one yourself
The diagonals of quadrilateral PQRS are perpendicular and equal. Must it be a square? If not, what extra condition makes it a square?
Show answer
No. A kite such as P(0, 4), Q(2, 1), R(0, 0), S(−2, 1) has PR = QS = 4 and PR ⟂ QS, but its diagonals do not bisect each other and it is not a square. If the diagonals also bisect each other, it is a square.
More questions like this
- 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 (see the figure). Why did we assume AB ≠ BC?
- On a piece of paper, draw and then cut out two copies of the same triangle. Keep one copy aside. On the other, mark the midpoint of each side. (You can do this by folding the paper to join two vertices at a time.) Draw the triangle made by the three midpoints and cut the paper along each of these lines. Now you have four smaller triangles. Compare them to each other and to the intact copy of the original triangle. What do you notice?
- Try to prove directly that the four smaller triangles are congruent. You will see that no congruence test applies, because many quantities are unknown.
- The segment joining the midpoints of two sides of a triangle is parallel to the third side and has half the length.
- Read the statement of the Midpoint Theorem again. The following questions seem natural to ask. What can we say about (1) a segment PQ (with P on side AB and Q on side AC) that is parallel to the third side and has half the length, and (2) the line that passes through the midpoint of one side and is parallel to another side? For now, experiment and see if you can make a guess about question (2).
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