Recall the following properties of a parallelogram that we proved last year.
- (a) The opposite sides of a parallelogram are equal.
- (b) The opposite angles of a parallelogram are equal.
- (c) The diagonals of a parallelogram bisect each other.
Step-by-step solution
Idea: In a parallelogram every diagonal is a transversal for two pairs of parallel lines, which gives equal alternate angles. Equal angles plus a common or equal side give congruent triangles.
(a) The opposite sides of a parallelogram are equal.
- Join AC. With AB ‖ DC and transversal AC: ∠CAB = ∠ACD (alternate angles). With AD ‖ BC and transversal AC: ∠ACB = ∠CAD (alternate angles).1 mark
- AC is common, so ∆ACD ≅ ∆CAB by ASA. Corresponding parts give AB = CD and AD = CB.1 mark
(b) The opposite angles of a parallelogram are equal.
- AD ‖ BC with transversal AB: ∠A + ∠B = 180° (co-interior angles). AB ‖ DC with transversal BC: ∠B + ∠C = 180°.1 mark
- So ∠A = 180° − ∠B = ∠C. In the same way ∠B + ∠C = 180° and ∠C + ∠D = 180° give ∠B = ∠D.½ mark
(c) The diagonals of a parallelogram bisect each other.
- Let the diagonals meet at E. In ∆AED and ∆CEB: AD = CB (from (a)); ∠EAD = ∠ECB and ∠EDA = ∠EBC (alternate angles, AD ‖ BC, with transversals AC and BD).1 mark
- So ∆AED ≅ ∆CEB (ASA), giving EA = EC and ED = EB: E is the midpoint of both diagonals.½ mark
Answer to write in the exam
(a)
Join AC. In ∆ACD and ∆CAB:
∠ACD = ∠CAB (alternate angles, AB ‖ DC)
AC = CA (common)
∠CAD = ∠ACB (alternate angles, AD ‖ BC)
∴ ∆ACD ≅ ∆CAB (ASA) ⇒ AB = CD and AD = CB (CPCT)
(b)
∠A + ∠B = 180° (co-interior angles, AD ‖ BC)
∠B + ∠C = 180° (co-interior angles, AB ‖ DC)
⇒ ∠A = ∠C
Similarly ∠C + ∠D = 180° ⇒ ∠B = ∠D
(c)
In ∆AED and ∆CEB:
∠EAD = ∠ECB, ∠EDA = ∠EBC (alternate angles, AD ‖ BC)
AD = CB (opposite sides of a parallelogram)
∴ ∆AED ≅ ∆CEB (ASA) ⇒ EA = EC, ED = EB (CPCT)
Common mistakes that cost marks
- Writing ∆ACD ≅ ∆ABC instead of ∆ACD ≅ ∆CAB. The order of letters must match the equal parts (A↔C, C↔A, D↔B).
- Calling ∠CAB and ∠ACD “corresponding angles”. They are alternate angles.
- In (c), using EA = EC as a reason. That is what has to be proved.
How this can come in the exam
In parallelogram PQRS, ∠P = (3x + 10)° and ∠Q = (2x + 20)°. The value of x is
- 10
- 30
- 50
- 34
Show answer
(B) 30
Adjacent angles are supplementary: 3x + 10 + 2x + 20 = 180 ⇒ 5x = 150 ⇒ x = 30.
The diagonals of parallelogram ABCD meet at O. If OA = 2y + 1 and OC = 3y − 4, find y and AC.
Show answer
Diagonals bisect each other: 2y + 1 = 3y − 4 ⇒ y = 5 (1 mark). OA = 11, so AC = 22 (1 mark).Try one yourself
In parallelogram ABCD, AB = 9 cm and the perimeter is 30 cm. Find BC.
Show answer
AB = CD = 9 and AD = BC. 2(9 + BC) = 30 ⇒ BC = 6 cm.
More questions like this
- If the converse of any of the three properties is true, it can be used as an alternate way to show that a quadrilateral is a parallelogram. To explore this, let us write the converses. Can you experiment and guess what the answers are?
- If the opposite sides of a quadrilateral are of equal length, then it is a parallelogram.
- If the opposite angles of a quadrilateral are equal, then it is a parallelogram.
- The angles of a quadrilateral add up to 360°. Therefore, if the opposite angles are equal, what can we say about adjacent angles? Is the converse of your answer true? Conclude that the result “if the opposite angles of a quadrilateral are equal, then it is a parallelogram” can also be stated as follows. “If each pair of adjacent angles in a quadrilateral ABCD …then ABCD is a parallelogram.” Fill in the blank.
- A quadrilateral whose diagonals bisect each other is a parallelogram.
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