Cubic EquationsQuadrilateral Properties
The Diagonals of a Cyclic Quadrilateral
Using the Law of Cosines, a formula for the length of the diagonals in term of the sides of the cyclic quadrilateral can be found. Using this formula we can also discover other relations pertaining to cyclic quadrilaterals.
In Figure 1, the sides of the cyclic quadrilateral are a, b, c, and d. Diagonals \( D_{1} \) and \( D_{2} \) have been drawn. By the Law of Cosines we have:
(i) \( D_{1}^{2} = a^{2} + d^{2} - 2ad\cos A \) and \( D_{1}^{2} = b^{2} + c^{2} - 2bc\cos C \)
Solving for \( \cos A \) and \( \cos C \) we obtain:
(ii) \( \cos A = \frac{a^{2} + d^{2} - D_{1}^{2}}{2ad} \) and \( \cos C = \frac{b^{2} + c^{2} - D_{1}^{2}}{2bc} \)
Adding these terms gives us \( \cos A + \cos C = \frac{2ad(b^{2} + c^{2} - D_{1}^{2}) + 2bc(a^{2}+d^{2}-D_{1}^{2})}{4abcd} \).
Because angles A and C are opposite angles of the quadrilateral, they are supplementary angles. Therefore and \( \cos A = -\cos C\) and \( \cos A + \cos C = 0 \). Therefore, we can solve for \(D_{1}\).
(iii) \( D_{1}^{2}(ad+bc) = ad(b^{2}+c^{2}) + bc(a^{2} + d^{2}) \)
\(D_{2}\) can be found by a similar method. Hence, we have the following:
The diagonals \( D_{1}^{2} \) and \( D_{2}^{2} \) of a cyclic quadrilateral in terms of the sides, have lengths of
\( D_{1}^{2} = \frac{ad(b^{2}+c^{2}) + bc(a^{2} + d^{2})}{ad+bc} \) and \( D_{2}^{2} = \frac{ab(c^{2}+d^{2}) + cd(a^{2} + b^{2})}{ab+cd} \).
After factoring, the diagonals can also be written as \( D_{1}^{2} = \frac{(bd+ac)(ab+cd)}{bc+ad} \) and \( D_{2}^{2} = \frac{(bc+ad)(ac+bd)}{ab+cd} \).
Unique Quadrilaterals
We can also prove that if a, b, c, and d are of any length where only of one of the values may equal zero, then the sides can be arranged in a way such that the vertices of the quadrilateral lie on a circle. Let’s first state our hypothesis as a theorem.
Theorem: For any given lengths of the sides of a quadrilateral where only one of the sides may have a length of zero, the sides can be arranged in a way such that its vertices lie on a circle.
Proof: Ptolemy’s theorem states that the product of the diagonals of a cyclic quadrilateral equals the sum of the product of the opposite sides. If we multiply \( D_{1}^{2} \) and \( D_{2}^{2} \), we should obtain \( (ac + bd)^{2} \) because we are dealing with the squares of the diagonals. Let a, b, c, and d be the lengths of the segments. Using these values as the sides, we find the length of the diagonals. Before multiplying we can simplify the formulas of the diagonals making it easier to carry out the multiplication.
Multiplying and factoring:
(i) \( D_{1}^{2} = \frac{adb^{2}+adc^{2}+bca^{2}+bcd^{2}}{bc+ad} = \frac{ab(ac+bd)+cd(bd+ac)}{bc+ad} \)
(ii) \( D_{1}^{2} = \frac{(bd+ac)(ab+cd)}{bc+ad} \)
By similar factoring:
(iii) \( D_{2}^{2} = \frac{(bc+ad)(ac+bd)}{ab+cd} \)
Now, we multiply the two diagonal lengths:
(iv) \( D_{1}^{2}D_{2}^{2} = \frac{(bc+ad)(ab+cd)(ac+bd)^{2}}{ab^{2}c+bc^{2}d+a^{2}bd+acd^{2}} \)
(v) \( D_{1}^{2}D_{2}^{2} = \frac{(bc+ad)(ab+cd)(ac+bd)^{2}}{ab(bc+ad)+cd(bc+ad)} \)
(vi) \( D_{1}^{2}D_{2}^{2} = \frac{(bc+ad)(ab+cd)(ac+bd)^{2}}{bc+ad)(ab+cd)} = (ac+bc)^{2} \)
Thus, for all values of a, b, c, and d where only one of the values may be zero, we can arrange the segments to form a cyclic quadrilateral!
Now that we have proved our theorem, the question is how many such cyclic quadrilaterals can be formed. Surprisingly we can form 3 distinct cyclic quadrilateral with the given lengths if no two sides are equal in length because the order of the sides affects the shape of the cyclic quadrilateral. The sides of the cyclic quadrilateral can be placed in order as follows: a, b, c, d; b, d, a, c; and c, a, b, d.
The 3 unique quadrilaterals are shown above with arbitrary lengths that are rounded to nearest 3 decimals. They all have the same circumradius. Their area would also be the same.
Cyclic Quadrilateral Area
The area of all three quadrilaterals is equal because their lengths are not altered and the area is given by Brahmagupta’s formula: \( K = \sqrt{(s-a)(s-b)(s-c)(s-d)} \), where K is the area, and s is the semiperimeter or \( s = \frac{1}{2}(a+b+c+d) \).
Consider a triangle as a quadrilateral with a side missing. All triangles have a circumcircle. The area of a triangle is given by the formula \( K = \sqrt{(s)(s-a)(s-b)(s-c)} \), where \( s = \frac{1}{2}(a+b+c) \).
Example
Calculating Diagonals Length
Let’s make up an example quadrilateral with sides we know. Our quadrilateral will have sides 6, 8, 5, 5√3. The radius will be 5 since the hypotenuse is 10. The quadrilateral forms two right triangles. Let’s calculate their diagonal lengths. We will start with diagonal BD, which is not touched by vertex A.
(i) \( D_{1}^{2} = \dfrac{(bd+ac)(ab+cd)}{bc+ad} \)
(ii) \( D_{1}^{2} = \dfrac{[(8)(5\sqrt{3})+(6)(5)][(6)(8)+(5)(5\sqrt{3})]}{(8)(5)+(6)(5\sqrt{3})} \)
(iii) \( D_{1}^{2} = \dfrac{[40\sqrt{3}+30][48+25\sqrt{3}]}{40+30\sqrt{3}} \)
(iv) \( D_{1}^{2} = \dfrac{[4\sqrt{3}+3][48+25\sqrt{3}]}{4+3\sqrt{3}} \)
(v) \( D_{1}^{2} = \dfrac{192\sqrt{3} + 300 + 144 + 75\sqrt{3}}{4+3\sqrt{3}} \)
(vi) \( D_{1}^{2} = \dfrac{444 + 267\sqrt{3}}{4+3\sqrt{3}} \cdot \dfrac{4-3\sqrt{3}}{4-3\sqrt{3}} \)
(vii) \( D_{1}^{2} = \dfrac{1776 - 1332\sqrt{3} + 1068\sqrt{3} - 2403}{16-27} \)
(viii) \( D_{1}^{2} = \dfrac{627 + 264\sqrt{3}}{11} = 57+24\sqrt{3} \approx 98.5628 \)
(ix) \( D_{1} = \sqrt{57+24\sqrt{3}} \approx 9.928 \)
Diagonal 1 length verifies with Geogebra. Diagonal 2 should be equal to 10. Let’s see if that evaluates correctly. Diagonal AC is not touched by B.
(i) \( D_{2}^{2} = \dfrac{(bc+ad)(ac+bd)}{ab+cd} \)
(ii) \( D_{2}^{2} = \dfrac{[(8)(5)+(6)(5\sqrt{3})][(6)(5)+(8)(5\sqrt{3})]}{(6)(8)+(5)(5\sqrt{3})} \)
(iii) \( D_{2}^{2} = \dfrac{[40+30\sqrt{3}][30+40\sqrt{3}]}{48+25\sqrt{3}} \)
(iv) \( D_{2}^{2} = \dfrac{4800 + 2500\sqrt{3}}{48+25\sqrt{3}} \)
(v) \( D_{2}^{2} = \dfrac{100(48 + 25\sqrt{3})}{48+25\sqrt{3}} = 100 \)
(vi) \( D_{2} = \sqrt{100} = 10 \)
Both diagonals check.
Calculating the Area
Let’s calculate the area of the quadrilateral 2 ways. It is made of 2 right triangles so the area is \( K = \frac{1}{2}(6)(8) + \frac{1}{2}{5}(5\sqrt{3}) = 24 + \frac{25}{2}\sqrt{3} \approx 45.6506 \).
The semiperimeter is \( \frac{1}{2}(6+8+5+5\sqrt{3}) = \frac{19}{2} + \frac{5}{2}\sqrt{3} \).
Using the formula \( K = \sqrt{(s-a)(s-b)(s-c)(s-d)} \), the area is:
(i) \( K = \sqrt{\left(\frac{19}{2} + \frac{5}{2}\sqrt{3} - 6\right)\left(\frac{19}{2} + \frac{5}{2}\sqrt{3} - 8\right)\left(\frac{19}{2} + \frac{5}{2}\sqrt{3} - 5\right)\left(\frac{19}{2} + \frac{5}{2}\sqrt{3} - 5\sqrt{3}\right)} \)
(ii) \( K = \sqrt{\left(\frac{7}{2} + \frac{5}{2}\sqrt{3}\right)\left(\frac{3}{2} + \frac{5}{2}\sqrt{3} \right)\left(\frac{9}{2} + \frac{5}{2}\sqrt{3}\right)\left(\frac{19}{2}-\frac{5}{2}\sqrt{3}\right)} \)
(iii) \( K = \frac{1}{4}\sqrt{\left(7 + 5\sqrt{3}\right)\left(3 + 5\sqrt{3} \right)\left(9 + 5\sqrt{3}\right)\left(19-5\sqrt{3}\right)} \)
(iv) \( K = \frac{1}{4}\sqrt{(96+50\sqrt{3})(96+50\sqrt{3})} \) (Multiplying the first 2 terms and last 2 terms inside the radical)
(v) \( K = 24 + \frac{25}{2}\sqrt{3} \)
The formula of the area checks.
The Cosine Values
Remember the relationship we obtained for cosine: \( \cos A = \frac{a^{2} + d^{2} - D_{1}^{2}}{2ad} \) and \( \cos C = \frac{b^{2} + c^{2} - D_{1}^{2}}{2bc} \). Since we know the lengths of the diagonals, we can solve for the cosine values in terms of the lengths.
(i) \( \cos A = \dfrac{a^{2} + d^{2} - \frac{ad(b^{2}+c^{2}) + bc(a^{2} + d^{2})}{ad+bc} }{2ad} \)
(ii) \( \cos A = \dfrac{a^{2}(ad+bc) + d^{2}(ad+bc) - [ad(b^{2}+c^{2}) + bc(a^{2} + d^{2})]}{2ad(ad+bc)} \)
(iii) \( \cos A = \dfrac{(ad)a^{2}+ (bc)a^{2} + (ad)d^{2} + (bc)d^{2} - (ad)b^{2} - (ad)c^{2} - (bc)a^{2} - (bc)d^{2}}{2ad(ad+bc)} \)
(iv) \( \cos A = \dfrac{(ad)a^{2} + (ad)d^{2} - (ad)b^{2} - (ad)c^{2}}{2bc(ad+bc)} \)
(v) \( \cos A = \dfrac{(ad)(a^{2} + d^{2} - b^{2} - c^{2})}{2bc(ad+bc)} \)
(vi) \( \cos A = \dfrac{(a^{2} + d^{2}) - (b^{2} + c^{2})}{2(ad+bc)} \)
By similar method, the other values are: \( \cos B = \dfrac{(b^{2} + a^{2}) - (c^{2} + d^{2})}{2(ba+cd)} \), \( \cos C = \dfrac{(c^{2} + b^{2}) - (d^{2} + a^{2})}{2(cb+da)} \), and \( \cos D = \dfrac{(d^{2} + c^{2}) - (a^{2} + b^{2})}{2(dc+ab)} \).
If we let one side equal 0, we return to the Law of Cosines. For example, \( \cos A = \dfrac{(a^{2} + d^{2}) - (b^{2} + c^{2})}{2(ad+bc)} \) becomes \( \cos A = \dfrac{(a^{2} + 0) - (b^{2} + c^{2})}{2(bc)} = \) \( 2bc\cos A = a^{2} - b^{2} - c^{2} = \) \( a^{2} = b^{2} + c^{2} - 2bc\cos A \).
Example
Continuing with our a = 6, b = 8, c = 5, and d = 5√3 quadrilateral, let’s find the cosine values of all 4 angles.
Angle A: \( \cos A = \dfrac{(a^{2} + d^{2}) - (b^{2} + c^{2})}{2(ad+bc)} = \) \( \dfrac{(6^{2} + (5\sqrt{3})^{2}) - (8^{2} + 5^{2})}{2((6)(5\sqrt{3})+(8)(5))} = \) \( \dfrac{(36 + 75) - (64 + 25)}{2(30\sqrt{3} + 40)} = \) \( \dfrac{111 - 89}{60\sqrt{3} + 80} = \) \( \dfrac{11}{10(4 + 3\sqrt{3})} \cdot \dfrac{4-3\sqrt{3}}{4-3\sqrt{3}} = \) \( \dfrac{11(4 - 3\sqrt{3})}{10(-11)} = \dfrac{-4+3\sqrt{3}}{10} \approx 0.1196\)
Angle B is a 90 degree angle. We should see a nice surprise.
Angle B: \( \cos B = \dfrac{(b^{2} + a^{2}) - (c^{2} + d^{2})}{2(ab+cd)} = \) \( \dfrac{(64 + 36) - (25 + 75)}{2((6)(8)+(8)(5\sqrt{3}))} = 0\)
Angle C is complementary to A. The sum of both is 0. Therefore, it should equal approximately -0.1196. Skipping some steps involving simple math:
Angle C: \( \cos C = \dfrac{(c^{2} + b^{2}) - (d^{2} + a^{2})}{2(bc+ad)} = \) \( \dfrac{(25 + 64) - (75 + 36)}{80+60\sqrt{3}} = \) \( \dfrac{-11}{10(4+3\sqrt{3})} = \) \( \dfrac{-11}{10(4+3\sqrt{3})} \cdot \dfrac{4-3\sqrt{3}}{4-3\sqrt{3}} = \) \( \dfrac{-11(4-3\sqrt{3})}{10(16-27)} = \) \( \dfrac{4-3\sqrt{3}}{10} \approx -0.1196 \)
Angle D is also 90 degrees, so cosine is 0.
Angle D: \( \cos D = \dfrac{(d^{2} + c^{2}) - (a^{2} + b^{2})}{2(ab+cd)} = \) \( \dfrac{(75 + 25) - (64 + 36)}{2((6)(8)+(8)(5\sqrt{3}))} = 0\)
Notice that \( \cos A = \cos(\angle DAC + \angle BAC) \). Let’s call these angles a and b. The sum angle identity of cosine is \( \cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b) \). Since these are right triangles, this evaluates to \( \frac{5\sqrt{3}}{10}\cdot \frac{6}{10} - \frac{5}{10} \cdot \frac{8}{10} = \) \( \frac{30\sqrt{3} - 40}{100} = \) \( \frac{3\sqrt{3} - 4}{10} \approx 0.1196 \). The cosine value verifies.
Diagonal Length of Any Quadrilateral
The above lengths were found without any trigonometry in the formulas because they were cyclic. To find the diagonal of any quadrilateral we must backup to where we substituted 0 for \( \cos A + \cos C \). Remember, to find \(D_{1}^{2} \), use angles whose vertices are not touched by \(D_{1}\). But to find \(D_{2}^{2} \) use the other pair of angles. The formulas of the diagonals of any quadrilateral are given as follows:
The diagonal lengths of any quadrilateral can be found by the following formulas:
\( D_{1}^{2} = \frac{ad(b^{2}+c^{2})+bc(a^{2}+d^{2}) - 2abcd(\cos A + \cos C}{ad+bc}) \) and \( D_{2}^{2} = \frac{ab(c^{2}+d^{2})+cd(a^{2}+b^{2}) - 2abcd(\cos B + \cos D)}{ab+cd} \)
Other Properties of Quadrilaterals
Ptolemy’s theorem states that \( D_{1}D_{2} = ac + bd \).
The circumradius of the cyclic quadrilateral is given by \( R = \frac{\sqrt{(ab+cd)(bc+ad)(ac+bd)}}{4K} \), where K is the area of the quadrilateral.
\( \frac{x}{y} = \frac{d}{b} \). Therefore, \( x = \frac{d(dD_{2}-bD_{1})}{d^{2}-b^{2}} \)
\( p = \frac{abD_{1}}{cD_{1}-aD_{2}} \) and \( q = \frac{abD_{2}}{cD_{1}-aD_{2}} \)