A quadrilateral with one pair of equal and parallel opposite sides is a parallelogram.
Step-by-step solution
To find: Prove that ABCD is a parallelogram
Idea: Turn the equal-and-parallel pair into equal triangles at the crossing point of the diagonals, so that the diagonals bisect each other; then use the diagonal test.
- AB ‖ DC, so ABCD is convex (all its vertices lie on two parallel lines with the figure between them) and its diagonals AC and BD meet at a point E.½ mark
- In ∆EAB and ∆ECD: ∠EAB = ∠ECD (alternate angles, AB ‖ DC, transversal AC); AB = CD (given); ∠EBA = ∠EDC (alternate angles, transversal BD). So ∆EAB ≅ ∆ECD (ASA).1½ marks
- Hence EA = EC and EB = ED: the diagonals bisect each other. A quadrilateral whose diagonals bisect each other is a parallelogram.1 mark
Answer to write in the exam
AB ‖ DC ⇒ ABCD convex ⇒ diagonals AC and BD meet at E
In ∆EAB and ∆ECD: ∠EAB = ∠ECD, ∠EBA = ∠EDC (alternate angles, AB ‖ DC); AB = CD (given)
∴ ∆EAB ≅ ∆ECD (ASA) ⇒ EA = EC, EB = ED (CPCT)
Diagonals bisect each other ⇒ ABCD is a parallelogram.
Common mistakes that cost marks
- Using only “AB = DC” or only “AB ‖ DC”. One pair must be both equal and parallel (a trapezium has a parallel pair; an isosceles trapezium also has an equal pair).
- Mixing the two pairs: AB ‖ DC with AD = BC is not enough (isosceles trapezium).
- Skipping why the diagonals meet: with AB ‖ DC the quadrilateral is convex.
How this can come in the exam
In quadrilateral PQRS, PQ ‖ SR and PQ = SR = 7 cm. If QR = 5 cm, then PS equals
- 7 cm
- 5 cm
- 12 cm
- cannot be found
Show answer
(B) 5 cm
One pair of opposite sides equal and parallel ⇒ PQRS is a parallelogram ⇒ PS = QR = 5 cm.
Try one yourself
ABCD is a parallelogram. M and N are the midpoints of AB and DC. Prove that AMCN is a parallelogram.
Show answer
AM = ½AB = ½DC = NC, and AM ‖ NC (parts of AB ‖ DC). One pair of opposite sides is equal and parallel, so AMCN is a parallelogram.
More questions like this
- Suppose in a quadrilateral ABCD we have AB ‖ DC and AB = DC. Let E be the intersection point of the diagonals AC and BD. (Why must the diagonals intersect?)
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- The diagonal AC of a parallelogram ABCD bisects ∠A. Show that it also bisects ∠C and that ABCD is a rhombus.
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- 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?
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