Law of cosines
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This article is about the law of cosines in Euclidean geometry. For the corresponding
theorem in spherical geometry, see law of cosines (spherical). For the cosine law of
optics, see Lambert's cosine law.
In trigonometry, the law of cosines (also known as Al-Kashi law or the cosine formula
or cosine rule) is a statement about a general triangle which relates the lengths of its
sides to the cosine of one of its angles. Using notation as in Fig. 1, the law of cosines
states that
or, equivalently:
Note that c is the side opposite of angle , and that a and b are the two sides enclosing .
All three of the identities above say the same thing; they are listed separately only
because in solving triangles with three given sides one may apply the identity three times
with the roles of the three sides permuted.The law of cosines generalizes the Pythagorean
theorem, which holds only in right triangles: if the angle is a right angle (of measure
90° or radians), then , and thus the law of cosines reduces to
which is the Pythagorean theorem. The law of cosines is useful for computing the third
side of a triangle when two sides and their enclosed angle are known, and in computing
the angles of a triangle if all three sides are known.
Stewart's theorem
Stewart's Theorem
In geometry, Stewart's theorem yields a relation between the lengths of the sides of a
triangle and the length of segment from a vertex to a point on the opposite side.
Let a, b, c be the sides of a triangle. Let p be a segment from A to a point on a dividing a
itself in x and y. Then
or alternatively:
Stewart's Theorem
In geometry, Stewart's theorem yields a relation between the lengths of the sides of a
triangle and the length of segment from a vertex to a point on the opposite side.
Let a, b, c be the sides of a triangle. Let p be a segment from A to a point on a dividing a
itself in x and y. Then
or alternatively:
Proof: Call the point where a and p meet P. We start applying the law of cosines to the
supplementary angles APB and APC.
Multiply the first by x the latter by y:
Now add the two equations:
and this is Stewart's theorem.
Apollonius' theorem
In elementary geometry, Apollonius' theorem is a theorem relating several elements in a
triangle. It states that given a triangle ABC, if D is any point on BC such that it divides
BC in the ratio n:m (or mBD = nDC), then
mAB2 + nAC2 = mBD2 + nDC2 + (m + n)AD2.
Special cases of the theorem
• When m = n( = 1), that is, AD is the median falling on BC, the theorem reduces to
• When in addition AB = AC, that is, the triangle is isosceles, the theorem reduces to the
Pythagorean theorem,
In simpler words, in any triangle , if is a median, then
To prove this theorem, let ' be a perpendicular dropped on from the point .
Then, in the right-angled triangles and , by Pythagoras' theorem, we
have
and
Adding equations (i) and (ii),
since
since is a right angle
And thus the theorem is proved.
Menelaus' theorem
Menelaus' theorem, case 1: line DEF passes inside triangle ABC
Menelaus' theorem, attributed to Menelaus of Alexandria, is a theorem about triangles
in plane geometry. Given points A, B, C that form triangle ABC, and points D, E, F that
lie on lines BC, AC, AB, then the theorem states that D, E, F are collinear if and only if:
In this equation, AB, etc., represent measurements of line segments that are allowed
negative values. For example, the fraction AF / FB must be defined as having positive
value only when line DEF intersects side AB, and similarly for the other fractions. There
is a long-running joke amongst mathematicians that, if this theorem is used to solve a
problem, then the 'wrong theorem' was being used (implying that Ceva's theorem should
have been used instead).
Proof
Menelaus' theorem, case 2: line DEF is entirely outside triangle ABC
This is one of many proofs for this theorem. The sign of the left-hand side of the
theorem's equation can be checked. Line DEF must intersect the sides of triangle ABC an
even number of times—either twice altogether, if it passes into the triangle and out again
(upper diagram), or not at all, if it misses the triangle (lower diagram). Hence there are an
odd number of negative terms, and the total product is negative.
Next, the magnitude can be checked. Construct line segments that connect line DEF
perpendicularly with vertices A, B, and C. With DEF as the base, let the altitudes of A, B,
and C be a, b, and c. By similar triangles, the absolute value of the left-hand side of the
theorem simplifies to:
Furthermore, we can prove by contradiction that if the theorem's equation holds, then D,
E, F must be collinear. Let there be a new point F' on AB different from F, and define the
measurements of AF, AF', and AB as n, n', and s. Suppose that F' also satisfies the
equation. Then the fractions have equal value:
which simplifies to n = n'. This proves that only one point on line AB can satisfy the
equation. With D and E fixed, if F satisfies the equation, then it must be the point
collinear with D and E. For reasons of symmetry, the same thing can be said about points
D and E.
Ptolemy used Menelaus' theorem as the basis for his spherical trigonometry in the
Almagest.
Ceva's theorem
Ceva's theorem, case 1: the three lines are concurrent at a point O inside ABC
Ceva's theorem, case 2: the three lines are concurrent at a point O outside ABC
Ceva's theorem is a well-known theorem in elementary geometry. Given a triangle ABC,
and points D, E, and F that lie on lines BC, CA, and AB respectively, the theorem states
that lines AD, BE and CF are concurrent if and only if
There is also an equivalent trigonometric form of Ceva's Theorem, that is, AD,BE,CF
concur if and only if
The theorem was proved by Giovanni Ceva in his 1678 work De lineis rectis, but it was
also proved much earlier by Al-Mu'taman ibn H d, an eleventh-century king of
Saragossa.
Associated with the figures are several terms derived from Ceva's name: cevian (the lines
AD, BE, CF are the cevians of O), cevian triangle (the triangle DEF is the cevian triangle
of O); cevian nest, anticevian triangle, Ceva conjugate. (Ceva is pronounced Chay'va;
cevian is pronounced chev'ian.)
Proof of the theorem
Suppose AD, BE and CF intersect at a point O. Because and have the
same height, we have
Similarly,
From this it follows that
Similarly,
and
Multiplying these three equations gives
as required. Conversely, suppose that the points D, E and F satisfy the above equality.
Let AD and BE intersect at O, and let CO intersect AB at F'. By the direction we have just
proven,
Comparing with the above equality, we obtain
Adding 1 to both sides and using AF' + F'B = AF + FB = AB, we obtain
Thus F'B = FB, so that F and F' coincide (recalling that the distances are directed).
Therefore AD, BE and CF = CF' intersect at O, and both implications are proven.
For the trigonometic form of the theorem, one approach is to view the three cevians,
concurrent at point O, as partitioning the triangle into three smaller triangles:
, , and .
Applying the law of sines to each triangle we get:
When we mutiply the three equations, the right side will equal 1. The six sines on the left
side, when rearranged, will yield the expression given in the theorem.
