The diagonal AC of a parallelogram ABCD bisects ∠A. Show that it also bisects ∠C and that ABCD is a rhombus.
Answer: Alternate angles give ∠ACD = ∠CAB and ∠ACB = ∠DAC. Since ∠CAB = ∠DAC, all four are equal, so ∠ACD = ∠ACB: AC bisects ∠C. In ∆ABC, ∠BAC = ∠BCA, so AB = BC; a parallelogram with two adjacent sides equal is a rhombus.
Step-by-step solution
Given: ABCD is a parallelogram; ∠DAC = ∠CAB (AC bisects ∠A)
To find: Prove that AC bisects ∠C and that ABCD is a rhombus
To find: Prove that AC bisects ∠C and that ABCD is a rhombus
Idea: Each half of ∠A has an alternate-angle partner at C. Then an isosceles triangle on the diagonal gives two adjacent sides equal.
- AB ‖ DC with transversal AC: ∠CAB = ∠ACD (alternate angles). AD ‖ BC with transversal AC: ∠DAC = ∠ACB (alternate angles).1 mark
- Given ∠DAC = ∠CAB, so ∠ACD = ∠CAB = ∠DAC = ∠ACB. Hence ∠ACD = ∠ACB: AC bisects ∠C.1 mark
- In ∆ABC, ∠BAC = ∠BCA, so the sides opposite them are equal: BC = AB. In a parallelogram AB = CD and BC = AD, so AB = BC = CD = DA: ABCD is a rhombus.1 mark
AC bisects ∠C, and AB = BC = CD = DA, so ABCD is a rhombus.
Answer to write in the exam
∠CAB = ∠ACD (alternate angles, AB ‖ DC)
∠DAC = ∠ACB (alternate angles, AD ‖ BC)
∠DAC = ∠CAB (given) ⇒ ∠ACD = ∠ACB ⇒ AC bisects ∠C
In ∆ABC: ∠BAC = ∠BCA ⇒ BC = AB (sides opposite equal angles)
AB = CD, BC = AD (parallelogram) ⇒ AB = BC = CD = DA
∴ ABCD is a rhombus.
Common mistakes that cost marks
- Pairing the wrong angles: ∠CAB pairs with ∠ACD (AB ‖ DC), and ∠DAC pairs with ∠ACB (AD ‖ BC).
- Saying “AB = BC because the angles at A are equal”. The equal angles needed are ∠BAC and ∠BCA, in the same triangle.
- Stopping at AB = BC without using the parallelogram to get all four sides equal.
How this can come in the exam
Short answer (2 marks)
In parallelogram PQRS, diagonal PR bisects ∠P. If PQ = 6 cm, find the perimeter of PQRS.
Show answer
As above, a parallelogram whose diagonal bisects an angle is a rhombus (alternate angles make ∆PQR isosceles with PQ = QR) (1 mark). All sides are 6 cm, so the perimeter is 24 cm (1 mark).Try one yourself
In rhombus ABCD, ∠ABC = 110°. Find ∠ACB.
Show answer
In ∆ABC, AB = BC, so ∠BAC = ∠BCA = (180° − 110°) ÷ 2 = 35°.
More questions like this
- The following questions examine converses of true properties. Answer them with Yes or No. If your answer is No, what extra condition can you add so that the answer becomes Yes?
- 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?
- 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.
All Quadrilaterals and parallelograms questions · All maths questions