Chapter 10
Quadrilaterals
10.1
Area formula
Consider a quadrilateral ABCD with sides
AB = a,
BC = b,
CD = c,
DA = d,
angles
6
DAB = α,
ABC = β,
6
6
BCD = γ,
6
CDA = δ,
and diagonals
AC = x,
BD = y.
B
a
A
~
b
d
x
y
Ñ
Å
D
C
c
Applying the cosine formula to triangles ABC and ADC, we have
x2 = a2 + b2 − 2ab cos β,
146
YIU: Euclidean Geometry
147
x2 = c2 + d2 − 2cd cos δ.
Eliminating x, we have
a2 + b2 − c2 − d2 = 2ab cos β − 2cd cos δ,
Denote by S the area of the quadrilateral. Clearly,
1
1
S = ab sin β + cd sin δ.
2
2
Combining these two equations, we have
=
=
=
=
=
16S 2 + (a2 + b2 − c2 − d2 )2
4(ab sin β + cd sin δ)2 + 4(ab cos β − cd cos δ)2
4(a2 b2 + c2 d2 ) − 8abcd(cos β cos δ − sin β sin δ)
4(a2 b2 + c2 d2 ) − 8abcd cos(β + δ)
β+δ
4(a2 b2 + c2 d2 ) − 8abcd[2 cos2
− 1]
2
β+δ
4(ab + cd)2 − 16abcd cos2
.
2
Consequently,
β+δ
2
= [2(ab + cd) + (a2 + b2 − c2 − d2 )][2(ab + cd) − (a2 + b2 − c2 − d2 )]
β +δ
−16abcd cos2
2
β+δ
2
= [(a + b) − (c − d)2 ][(c + d)2 − (a − b)2 ] − 16abcd cos2
2
= (a + b + c − d)(a + b − c + d)(c + d + a − b)(c + d − a + b)
β +δ
.
−16abcd cos2
2
16S 2 = 4(ab + cd)2 − (a2 + b2 − c2 − d2 )2 − 16abcd cos2
Writing
2s := a + b + c + d,
we reorganize this as
S 2 = (s − a)(s − b)(s − c)(s − d) − abcd cos2
β+δ
.
2
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10.1.1
148
Cyclic quadrilateral
If the quadrilateral is cyclic, then β + δ = 180◦ , and cos β+δ
2 = 0. The area
formula becomes
S=
q
(s − a)(s − b)(s − c)(s − d),
where s = 12 (a + b + c + d).
Exercise
1. If the lengths of the sides of a quadrilateral are fixed, its area is greatest
when the quadrilateral is cyclic.
2. Show that the Heron formula for the area of a triangle is a special case
of this formula.
10.2
Ptolemy’s Theorem
Suppose the quadrilateral ABCD is cyclic. Then, β + δ = 180◦ , and cos β =
− cos δ. It follows that
a2 + b2 − x2 c2 + d2 − x2
+
= 0,
2ab
2cd
and
(ac + bd)(ad + bc)
.
ab + cd
Similarly, the other diagonal y is given by
x2 =
y2 =
(ab + cd)(ac + bd)
.
(ad + bc)
From these, we obtain
xy = ac + bd.
This is Ptolemy’s Theorem. We give a synthetic proof of the theorem and
its converse.
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10.2.1
149
Ptolemy’s Theorem
A convex quadrilateral ABCD is cyclic if and only if
AB · CD + AD · BC = AC · BD.
Proof. (Necessity) Assume, without loss of generality, that 6 BAD > 6 ABD.
Choose a point P on the diagonal BD such that 6 BAP = 6 CAD. Triangles BAP and CAD are similar, since 6 ABP = 6 ACD. It follows that
AB : AC = BP : CD, and
AB · CD = AC · BP.
Now, triangles ABC and AP D are also similar, since 6 BAC = 6 BAP +
6 P AC = 6 DAC + 6 P AC = 6 P AD, and 6 ACB = 6 ADP . It follows that
AC : BC = AD : P D, and
BC · AD = AC · P D.
Combining the two equations, we have
A
AB · CD + BC · AD = AC(BP + P D) = AC · BD.
A
D
D
O
P'
P
B
C
B
C
(Sufficiency). Let ABCD be a quadrilateral satisfying (**). Locate a
point P 0 such that 6 BAP 0 = 6 CAD and 6 ABP 0 = 6 ACD. Then the
triangles ABP and ACD are similar. It follows that
AB : AP 0 : BP 0 = AC : AD : CD.
From this we conclude that
(i) AB · CD = AC · BP 0 , and
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150
(ii) triangles ABC and AP 0 D are similar since
AB : AC = AP 0 : AD.
Consequently, AC : BC = AD : P 0 D, and
6
BAC =
6
P 0 AD and
AD · BC = AC · P 0 D.
Combining the two equations,
AC(BP 0 + P 0 D) = AB cot CD + AD · BC = AC · BD.
It follows that BP 0 + P 0 D = BC, and the point P 0 lies on diagonal BD.
From this, 6 ABD = 6 ABP 0 = 6 ACD, and the points A, B, C, D are
concyclic.
Exercise
1. Let P be a point on the minor arc BC of the circumcircle of an equilateral triangle ABC. Show that AP = BP + CP .
A
C
B
P
2. P is a point on the incircle of an equilateral triangle ABC. Show that
AP 2 + BP 2 + CP 2 is constant. 1
1
If each side of the equilateral triangle has length 2a, then AP 2 + BP 2 + CP 2 = 5a2 .