In a quadrilateral ABCD, suppose AB ‖ DC. Can ABCD be non-convex? What if we instead assume AB = CD? What if we instead assume ∠A = ∠C?
Step-by-step solution
Idea: A quadrilateral is non-convex when one internal angle is more than 180°. With AB ‖ DC every vertex sits on one of two parallel lines with the whole figure between them, so no angle can exceed 180°. For the other two conditions, one example (a counterexample) is enough to say “yes”.
- AB ‖ DC. A and B lie on one line, C and D on a parallel line. The sides AD and BC cross the gap between the lines, so the whole quadrilateral lies in the strip between the two parallel lines.1 mark
- Each vertex is on an edge of this strip, and the quadrilateral lies on one side of that edge line. So the internal angle at every vertex is less than 180°. (Also, with AD as transversal, ∠A + ∠D = 180°, and with BC as transversal, ∠B + ∠C = 180°.) Hence ABCD cannot be non-convex: it is always convex.1 mark
- AB = CD. Take A(0, 0), B(6, 0), C(3, 1), D(3, 7). AB = 6 and CD = 7 − 1 = 6. C lies inside triangle ABD, so the internal angle at C is reflex (about 252°). So ABCD can be non-convex.1 mark
- ∠A = ∠C. Take the arrowhead A(−4, 0), B(0, 1.5), C(4, 0), D(0, 7). It is symmetric about the line BD, so ∠A = ∠C, while B lies inside triangle ACD, giving a reflex angle at B. So ABCD can be non-convex. (The dent cannot be at A or C: then both A and C would be reflex, adding to more than 360°.)1 mark
Check: For the first example: ∠A ≈ 66.8°, ∠B ≈ 18.4°, ∠C ≈ 251.6°, ∠D ≈ 23.2°; sum = 360° ✓ and ∠C > 180°. For the arrowhead: ∠A = ∠C ≈ 39.7°, ∠D ≈ 59.5°, reflex ∠B ≈ 221.1°; sum = 360° ✓.
Answer to write in the exam
AB ‖ DC ⇒ A, B on one line and C, D on a parallel line; ABCD lies in the strip between them
⇒ every internal angle < 180° (∠A + ∠D = 180°, ∠B + ∠C = 180°, co-interior angles) ⇒ ABCD is convex
∴ With AB ‖ DC, ABCD cannot be non-convex.
AB = CD: A(0, 0), B(6, 0), C(3, 1), D(3, 7); AB = CD = 6, C inside ∆ABD ⇒ reflex ∠C ⇒ non-convex: yes, possible
∠A = ∠C: A(−4, 0), B(0, 1.5), C(4, 0), D(0, 7); symmetric about BD ⇒ ∠A = ∠C, B inside ∆ACD ⇒ non-convex: yes, possible
Common mistakes that cost marks
- Answering “no” for AB = CD or ∠A = ∠C after trying only nice shapes. A single non-convex example proves “yes, it can”.
- For AB ‖ DC, forgetting to give a reason; the strip between the parallel lines (or the co-interior angles) is the reason.
- Thinking a trapezium can have a dent. With one pair of sides parallel it is always convex.
How this can come in the exam
Assertion (A): A quadrilateral with one pair of opposite sides parallel is always convex.
Reason (R): Co-interior angles between two parallel lines add up to 180°.
- 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
(A) Both A and R are true, and R is the correct explanation of A.
With AB ‖ DC, ∠A + ∠D = 180° and ∠B + ∠C = 180°, so every angle is less than 180°: the quadrilateral is convex. R is the reason.
Which condition on quadrilateral ABCD forces it to be convex?
- AB = CD
- ∠A = ∠C
- AD ‖ BC
- AC = BD
Show answer
(C) AD ‖ BC
One pair of opposite sides parallel keeps every angle below 180°. The other conditions all allow a dent.
Try one yourself
Give coordinates of a non-convex quadrilateral PQRS with PQ = RS.
Show answer
For example P(0, 0), Q(8, 0), R(4, 2), S(4, 10): PQ = 8, RS = 8, and R lies inside triangle PQS, so the angle at R is reflex.
More questions like this
- Consider three non-collinear points A, B, C and draw the lines AB, BC, CA. For every possible location of point D in the plane outside these lines, decide if ABCD is self-intersecting, non-convex, or convex. (Hint: the three lines divide the plane into 7 regions.)
- Can a quadrilateral be both self-intersecting and non-planar?
- To test if a given quadrilateral is a parallelogram, do we have to check that the opposite sides are parallel? Are there other ways to test this?
- Recall the following properties of a parallelogram that we proved last year.
- 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?
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