Quadrilateral ABCD satisfies 2 A B ¯ = A C ¯ , B C ¯ = 3 ¯ , B D ¯ = D C ¯ and &lt; B A C = 60

Question

Quadrilateral ABCD satisfies             2      A      B        ¯    =            A      C        ¯   ,              B      C        ¯    =            3        ¯   ,             B      D        ¯    =            D      C        ¯   and   &amp;lt;  B  A  C  =  60

pyphekam · Accepted Answer

Step 1First note that since   &amp;lt;  B  A  C  =  60 and   2  A  B  =  A  C it tells us   Δ  A  B  C is congruent to the general 30, 60, 90 triangle (by SAS). This tells us that   &amp;lt;  A  B  C  =  90.Using the fact that the hypotenuse of an inscribed right triangle is a diameter then we have that   r  =            A      C        2   but as you stated   A  C  =  2 so   r  =  1Correctly by using laws of cosines   A  C  =  2Note that we know   &amp;lt;  B  A  C  =  60 and   &amp;lt;  B  D  C lies on the same arc as   &amp;lt;  B  A  C (arc BC) then we know these angle equal so   &amp;lt;  B  D  C  =  60Now we know that   &amp;lt;  B  D  C  =  60 we can see that   Δ  B  D  C is an congruent to the general equilateral triangle (by SAS since   B  D  =  D  C ). This now tells us   B  D  =  D  C  =  B  C  =      3   so the area of   Δ  B  D  C  =      1    2        3        3    s  i  n  (  60  )  =            3              3              4  We have already worked out that       |    C  A      |    =  2 and       |    D  C      |    =      3   . Then since   &amp;lt;  A  B  C  =  90 and   &amp;lt;  D  B  C  =  60 we have that   &amp;lt;  A  B  D  =  30 and therefore since angles in the triangle   Δ  A  B  X (where X the intersection of lines AC and BD) is add to 180, lines BD and AC meet at 90 degrees. Hence ΔBCX is congruent to   Δ  D  C  X (by AAS) which tells us that the angle between CA and DC is equal to   &amp;lt;  A  C  B  =  30 . Hence the dot product of CA and DC is   −  2      3    cos  ⁡      30    =  −  3If you want a short extension then show that quadrilateral ABCD is a kite

gvaldytist · Accepted Answer

Step 1Let   x  =            A      B        ¯    , then             A      C        ¯    =  2  x, and we are given that             B      C        ¯    =      3   and that   ∠  B  A  C  =      60    ∘  Applying the law of cosines to   △  A  B  C , we get  (      3        )    2    =      x    2    +  (  2  x      )    2    −  2  x  (  2  x  )  (      1    2    )Hence,  3  =      x    2    +  4      x    2    −  2      x    2    =  3      x    2  From which,   x  =  1 . Therefore,             A      B        ¯    =  1 and             A      C        ¯    =  2Now using coordinate geometry, we can set vertex   A  =  (  0  ,  0  ) and   B  =  (  1  ,  0  ) , then   C  =  (  1  ,      3    ) (Assuming C is in the upper half plane).Next, we find the circle on which A, B, C lies. The center is the intersection of the perpendicular bisector of AB, which is the line   x  =            1      2      , and the perpendicular bisector of BC, which is the line   y  =                    3            2      Hence the center of the circumcircle of   △  A  B  C is   O  =  (            1      2        ,                    3            2        )The radius of this circumcircle is   R  =            O      A        ¯    =  1Since point D lies on the cicrcumcircle and on perpendicular bisector of BC (because it is equidistant from B and C), then we D can be      D    1    =  O  +  (  −  R  ,  0  )  =  (  −            1      2        ,                    3            2        )    (  1  )or      D    2    =  O  +  (  R  ,  0  )  =  (            3      2        ,                    3            2        )    (  2  )However, from the cyclic order of the vertices,       D    2   is rejected. Hence   D  =      D    1   .Now we can compute the area.The area of   A  B  C  D  =  [  A  B  C  ]  +  [  A  C      D    1    ]  =      1    2        (          3        +                            3                2              −          3        (    −                  1        2              )    )    =      3  Also, we have                     C        A            →        =  A  −  C  =  (  −  1  ,  −      3    ) , and                     D        C            →        =  C  −  D  =  (            3      2        ,                    3            2        )So                     C        A            →        ⋅                    D        C            →        =  −            3      2        −            3      2        =  −  3

Quadrilateral ABCD satisfies <mover> 2 A B </mrow> <mo acce

Answered question

Answer & Explanation

New Questions in Elementary geometry