Relations Among Length and angles in General Parallelogram

This article aims to derive some relations between lengths and angles in general parallelograms (GPs), i.e., parallelograms that are neither rectangles nor rhombi. We shall use Figure 1 for reference. In parallelogram ABCD, AB > AD, and ­DAB is acute. Certain angles have been named p, q,r, s and t.


 Once the lengths AB, AD and ­Angle A are given, a unique parallelogram is defined. The lengths of the diagonals and angles p, q,r, s and t get fixed. (Do you see why?) We shall obtain expressions relating these to the given sides AB, AD and angle  ­A. 

The cosine rule of triangles can be used to obtain the following relations between the sides and the diagonals.

AC2 = AB2+ AD2 + 2AB.AD cos A

 BD2= AB2 + AD2 − 2AB.AD cos A.

These relations can be reworked to give

2(AB2 + AD2 ) = AC2 + BD2.

 This relation is equivalent to Apollonius’s theorem. It enables one to obtain one of the lengths given the other three.

The above pair of equations also lead to

 cos ­DAB = (AC2 − BD2 )/4AB.AD,

enabling one to obtain ­DAB from the four lengths.

 Using the sine rule in triangles ACD, ABC, ABD and BCD in turn, we obtain the following relations between angles p, q,r, s and ­DAB. Note that sin D = sin(180 − A) = sin A.

                             sin p/sin ­DAB = sin p/sin ­ADC = sin ­CAD/sin ­ADC = CD/AC = AB/AC, 

                             sin q/sin ­DAB = sin q/sin­CBA = sin ­CAB/sin ­ABC = BC/AC = AD/AC,

                             sin r/sin ­DAB = sin ­ABD/sin ­BAD = AD/BD,

                             sin s/sin ­DAB = sin s/sin C = sin ­CBD/sin ­BCD = CD/BD = AB/BD.

 To obtain a relation between angle t and ­A, we note the following.

 Area of parallelogram ABCD = AB.AD sin A = 1 2AC.BD sin t, giving

                                                      sin t/sin A = 2AB.AD/AC.BD

To summarise sin

                                               p/sin A = AB/AC

                                               sin q/sin A = AD/AC

                                                sin r/sin A = AD/BD

                                                sins/sin A = AB/BD

                                                sin t/sin A = 2AB. AD/AC.BD

Observe the neat pattern in the above set of equations. Several inferences can be made from the above.

 (i) sin p · sin r = sin q · sin s

 (ii) Angle inequalities: q < p <  s since AD/ AC < AB /AC < AB /BD , and q < r < s since AD/ AC < AD /BD < AB /BD .

 (iii) In the general case, angles p, q,r, s are distinct. Can any two of these be equal in a General Parallelogram (GP)?

If p = q, then ­DAC = ­ACD. Consequently AD = DC and the figure becomes a rhombus. A similar situation arises if r = s. If q = r or p = s, then the diagonals are of the same length, making the figure a rectangle. If q = s, then by the above inequalities, p = q = r = s = 45◦ and the figure turns out to be a square.

We now examine the pair p,r. If p = r, then we have AB/AC = AD/BD or AB/AD = AC/BD. That is, the ratio of the (adjacent) sides equals the ratio of the diagonals. This is true of a square. We ask if this is possible in a GP.

Taking AB = 1, AD = x < 1, AC = y and BD = xy, we have

2(1 + x 2) = y2+ x 2y 2= y2(1 + x 2), leading to

y = √2, i.e., the longer diagonal must be √2 times the longer side for angles p and r to be equal. The ratio of AD to AB can be any value between (√2 − 1) and 1. To see why, consider triangle ABD and apply the triangle inequality to its sides, which are 1, x, x √2.

If p = r then t = ­A. (This follows from the relations ­A = p + q and t = q + r.) That is, the angle between the sides equals the angle between the diagonals. This is again true of a square but is also true of the kind of GP described above.


Every parallelogram has its dual figure, whose sides are parallel to the diagonals of the former and vice versa. (Note that the concern here is only the shape of the figure, and not the scale. We can scale up the dual figure by any desired scale factor.) An easy way to obtain the dual of any parallelogram is to connect the mid-points of the sides sequentially, as indicated in


A rectangle and a rhombus where the angle between the sides in one is equal to the angle between the diagonals in the other are duals of each other. The dual of a square is another square. A GP of the type described in the previous paragraph, i.e., where the longer diagonal is √2 times the longer side, is its own dual.

To summarise, a member of this class of GPs shares the following properties with a square:

Ratio of sides equals ratio of diagonals. The longer diagonal is √2 times the longer side, while the shorter diagonal is √2 times the shorter side. The angle between the diagonals equals angle between sides It is its own dual.

We shall call a GP of this type a ‘PSEUDOSQUARE’.

Editor’s note. In the March 2017 issue of AtRiA, author Michael de Villiers describes a ‘Golden rectangle’ (pp. 64-69). In this figure the (adjacent) sides as well as the diagonals are in the ratio ϕ : 1, ϕ being the Golden ratio. Interestingly, the acute angle of the figure turns out to be 60◦. Such a golden rectangle is a special case of a pseudosquare, as defined above, where the longer diagonal is √2 times the longer side. It can be shown that if the acute angle of a pseudosquare is 60◦, then the sides and diagonals are in the golden ratio, and vice versa.


With reference to Figure 3, ABCD is a parallelogram with AB = 1 unit; A is placed at the origin and B on the positive X-axis. Angle q is an arbitrary angle less than 45◦. AC = √2 units. Point D completes the parallelogram. The coordinates of A, B, C, D are as indicated in the figure.

We first show that the ratio of sides equals the ratio of diagonals.

AB = 1. AC = √2.

AD2= (√2 cos q − 1)22 + (√2 sin q) 2, which simplifies to 3 − 2 √2 cos q, hence

AD = √[3 − 2 √2 cos q].

BD2 = (√2 cos q − 2)2+ (√2 sin q)2, which simplifies to 2(3 − 2 √2 cos q), hence

BD = √[2(3 − 2 √2 cos q)].Hence BD/AD = √2 = AC/AB.

Next, we show that the angle between the sides equals the angle between the diagonals.

Slope of AD = √2 sin q/√2 cos q−1 , while

Slope of AB = 0.

tan value of the angle between the lines is √2 sin q/√2 cos q−1 = 2 sin q/2 cos q−√2

 Slope of BD = √2 sin q/√2 cos q−2 , while

slope of AC = √2 sin q/√2 cos q .

tan of the angle between the lines is again 2 sin q /2cos q-√2

A.RAMACHANDRAN has had a longstanding interest in the teaching of mathematics and science. He studied physical science and mathematics at the undergraduate level, and shifted to life science at the postgraduate level. He taught science, mathematics and geography to middle school students at Rishi Valley School for two decades. His other interests include the English language and Indian music. He may be contacted at

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