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?)
Step-by-step solution
Idea: Diagonals fail to meet only in a non-convex quadrilateral. A pair of parallel opposite sides rules out a dent.
- A and B lie on one line, D and C on a parallel line, and the whole quadrilateral lies in the strip between these lines. At each vertex the figure stays on one side of the line through that vertex, so every internal angle is less than 180°: ABCD is convex. (Co-interior angles: ∠A + ∠D = 180° and ∠B + ∠C = 180°.)1 mark
- In a convex quadrilateral, each diagonal splits it into two triangles and the other two vertices lie on opposite sides of it, so the diagonals AC and BD cross at a point E inside.1 mark
Answer to write in the exam
AB ‖ DC ⇒ ∠A + ∠D = 180°, ∠B + ∠C = 180° (co-interior angles) ⇒ every angle < 180°
⇒ ABCD is convex
∴ Diagonals AC and BD intersect (at a point E inside ABCD).
Common mistakes that cost marks
- Assuming every quadrilateral’s diagonals meet. In a non-convex quadrilateral they do not.
- Using AB = DC as the reason. It is the parallel sides that rule out a dent.
How this can come in the exam
Assertion (A): The diagonals of every trapezium intersect.
Reason (R): A trapezium is a convex quadrilateral.
- 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.
A trapezium has a pair of parallel sides, so it is convex (R true), and the diagonals of a convex quadrilateral intersect (A true). R explains A.
Try one yourself
In quadrilateral ABCD, AD ‖ BC. Can vertex A lie inside triangle BCD?
Show answer
No. AD ‖ BC makes ABCD convex, so no vertex lies inside the triangle formed by the other three.
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