The diagonals AC and BD of a parallelogram ABCD intersect at O. A line through O meets AB and CD at points P and Q respectively. Show that O is the midpoint of PQ. (Multiple proofs are possible. Which is the simplest?)
Step-by-step solution
To find: Prove that OP = OQ
Idea: O is already the midpoint of AC. Two triangles on either side of O, one with side OA and one with side OC, are congruent because of the parallel sides.
- The diagonals of a parallelogram bisect each other, so OA = OC.½ mark
- AB ‖ DC with transversal AC: ∠OAP = ∠OCQ (alternate angles). ∠AOP = ∠COQ (vertically opposite angles).1 mark
- So ∆OAP ≅ ∆OCQ (ASA), giving OP = OQ: O is the midpoint of PQ.1 mark
- Which proof is simplest? This one: a single congruence. (Another proof: a half-turn about O swaps A with C and B with D, so it carries line AB onto line CD; P goes to the point of CD on line PO, which is Q, so OP = OQ. It is neat but needs more explanation.)½ mark
Check: A(0, 0), B(6, 0), C(8, 4), D(2, 4): O = (4, 2). The line through O and P(3, 0) meets DC (y = 4) at Q(5, 4); midpoint of PQ = (4, 2) = O ✓.
Answer to write in the exam
OA = OC (diagonals of a parallelogram bisect each other)
∠OAP = ∠OCQ (alternate angles, AB ‖ DC); ∠AOP = ∠COQ (vertically opposite angles)
∴ ∆OAP ≅ ∆OCQ (ASA) ⇒ OP = OQ (CPCT)
∴ O is the midpoint of PQ. (Simplest proof: this single congruence.)
Common mistakes that cost marks
- Using OB = OD with angles at A and C; the side must sit between (or match) the chosen angles. OA = OC fits ∠OAP and ∠OCQ.
- Calling ∠OAP and ∠OCQ corresponding angles; they are alternate angles.
- Assuming P and Q are midpoints of AB and CD. The line through O can have any direction.
How this can come in the exam
In parallelogram ABCD, the diagonals meet at O. A line through O meets AD at X and BC at Y. If OX = 3.5 cm, find XY.
Show answer
∆OAX ≅ ∆OCY (ASA: OA = OC, alternate angles, vertically opposite angles), so OY = OX = 3.5 cm (1 mark). XY = 7 cm (1 mark).Try one yourself
A line through the centre O of a rectangular table top meets two opposite edges at P and Q, with PQ = 1.3 m. Find OP.
Show answer
A rectangle is a parallelogram, so O is the midpoint of PQ: OP = 0.65 m.
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- (i) Suppose ABCD is a parallelogram and M, N are midpoints of AB and CD respectively. Show that segments DM and BN trisect segment AC.
(ii) Use part (i) to find a procedure to trisect any given segment PQ. Find two other ways to trisect PQ, one using the fact that a line parallel to one side of a triangle divides the other two sides in the same ratio and a third using the Centroid Theorem. - Is the Midpoint Theorem for Quadrilaterals true when the quadrilateral is non-convex? How about when it is self-intersecting? Experiment and check. The pictures may look strange, but the theorem appears to be true.
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