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?
Step-by-step solution
Idea: In the triangle formed by side AB and the two bisectors from A and B, the angles at A and B are ½∠A and ½∠B, which add up to 90°. So the third angle, at P, is 90°.
- Let P be where the bisectors of ∠A and ∠B meet. ∠A + ∠B = 180° (adjacent angles of a parallelogram), so ∠PAB + ∠PBA = ½∠A + ½∠B = 90°.1 mark
- In ∆APB the angles add to 180°, so ∠APB = 180° − 90° = 90°. The angle of PQRS at P lies between the same two bisector lines, so it is 90°.1 mark
- The same argument at Q (bisectors of B and C), R (C and D) and S (D and A) gives 90° each. A quadrilateral with four right angles is a rectangle: PQRS is a rectangle.1 mark
- Why AB ≠ BC? If AB = BC, ABCD is a rhombus and each diagonal bisects the angles at its ends. Then the bisectors of A and C are both the diagonal AC, and those of B and D are both BD, so all four intersection points are the centre: P = Q = R = S and no rectangle is formed.1 mark
Check: Example: A(0, 0), B(5, 0), D(2, 3), C(7, 3). Computing the four crossing points gives a quadrilateral whose sides meet at 90° at each vertex, with sides about 0.66 and 1.23, not equal (so a rectangle that is not a square).
Answer to write in the exam
∠A + ∠B = 180° (adjacent angles of a parallelogram)
In ∆APB: ∠PAB + ∠PBA = ½(∠A + ∠B) = 90° ⇒ ∠APB = 90°
Similarly ∠Q = ∠R = ∠S = 90°
∴ PQRS has four right angles ⇒ PQRS is a rectangle.
If AB = BC, ABCD is a rhombus; bisectors of ∠A, ∠C lie along AC and of ∠B, ∠D along BD ⇒ P = Q = R = S (the centre) ⇒ no rectangle. Hence AB ≠ BC.
Common mistakes that cost marks
- Proving only one right angle. All four angles of PQRS must be shown to be 90° (or one right angle plus a reason it is a parallelogram).
- Writing ∠A + ∠B = 360°. Adjacent angles of a parallelogram add to 180°.
- Saying AB ≠ BC is needed “so that the figure is not a square”. The real reason is that in a rhombus the four points collapse into one.
How this can come in the exam
In parallelogram ABCD, the bisectors of ∠A and ∠B meet at P. Then ∠APB is
- 45°
- 60°
- 90°
- it depends on ∠A
Show answer
(C) 90°
½∠A + ½∠B = ½ × 180° = 90°, so ∠APB = 180° − 90° = 90°, whatever ∠A is.
Try one yourself
In parallelogram ABCD, ∠A = 70°. The bisectors of ∠A and ∠D meet at S. Find ∠ASD and ∠SAD.
Show answer
∠SAD = 35°, ∠SDA = ½ × 110° = 55°, so ∠ASD = 180° − 90° = 90°.
More questions like this
- On a piece of paper, draw and then cut out two copies of the same triangle. Keep one copy aside. On the other, mark the midpoint of each side. (You can do this by folding the paper to join two vertices at a time.) Draw the triangle made by the three midpoints and cut the paper along each of these lines. Now you have four smaller triangles. Compare them to each other and to the intact copy of the original triangle. What do you notice?
- Try to prove directly that the four smaller triangles are congruent. You will see that no congruence test applies, because many quantities are unknown.
- The segment joining the midpoints of two sides of a triangle is parallel to the third side and has half the length.
- Read the statement of the Midpoint Theorem again. The following questions seem natural to ask. What can we say about (1) a segment PQ (with P on side AB and Q on side AC) that is parallel to the third side and has half the length, and (2) the line that passes through the midpoint of one side and is parallel to another side? For now, experiment and see if you can make a guess about question (2).
- The line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side (and the resulting segment has half the length of the parallel side).
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