Page 9 - MATINF Nr. 6
P. 9

Two Identities and their Consequences                                                           9























                                     Fig. 3: A general convex quadrilateral ABCD.


                                           2
                                        2
                                  2
                                     2
                Yielding cos α =  a +d −b −c +2bc cos γ  . Now, making use of the half angle formula for cosine,
                                         2ad
                                                       2
                                       2
                                                            2
                                            2
                                α    a + d + 2ad − b − c + 2bc cos γ
                           cos 2   =                                                                      (7)
                                2                    4ad
                                            2
                                                       2
                                                            2
                                       2
                                     a + d + 2ad − b − c + 2bc(1 − 2 sin    2 γ )
                                   =                                          2
                                                         4ad
                                                       2
                                            2
                                     (a + d) − (b − c) − 4bc sin 2 γ
                                   =                               2
                                                   4ad
                                     (a + d + b − c)(a + d − b + c) − 4bc sin 2 γ
                                   =                                          2
                                                         4ad
                                                          ‹                    ‹         2 γ
                                      1   a + b + c + d        a + b + c + d         bc sin
                                   =                    − c                  − b −          2
                                     ad         2                    2                  ad
                                     (s − b)(s − c) − bc sin 2 γ
                                   =                         2  .
                                                ad

                As in Theorem 1, the other formula can be obtained similarly by replacing cos 2 α  by 1−sin 2 α
                                                                                               2            2
            in (7).
                                                                                             ◦
                In the case of a cyclic convex quadrilateral, you get (1) by replacing  γ 2  by 90 −  α  in (5) and
                                                                                                 2
                                   ◦
            (6), since α + γ = 180 .
            Theorem 6 (Bretschneider). Given a general quadrilateral with sides a, b, c and d. If α and γ
            are two opposite angles, then the area is given by the formula
                                     É
                                                                                  α + γ

                               ∆ 3 =    (s − a)(s − b)(s − c)(s − d) − abcd cos 2        .
                                                                                    2
            Proof. Multiplying (5) and (6) we get
                              α           γ          γ           α

                      ad sin 2  + bc cos 2    bc sin 2  + ad cos 2   = (s − a)(s − b)(s − c)(s − d).
                              2           2          2           2
                Expanding, factorizing, completing the squares and keeping in mind some well-known
            trigonometric identities,

                                 α + γ     ad sin α   bc sin γ  ‹ 2
                       abcd cos 2         +           +            = (s − a)(s − b)(s − c)(s − d).
                                    2            2          2
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