Prove that angle bisectors of a triangle are concurrent using vectors. Also, find the position vector of the point of concurrency in terms of position vectors of the vertices.I solved this without using vectors to get some idea. I am not sure how to prove it using vectors. I don't want to use vector equations for straight lines and then find the point of concurrency. That's like solving using coordinate geometry.Let a → , b → , c → represent the sides A , B , C respectively.The angle bisectors are along a → | a → | + b → | b → | , b → | b → | + c → | c → | , c → | c → | + a → | a → | Let the sides A B , B C , C A be x , y , z. Let A D be one of the angular bisector. B D C D = x z Hence D = x c → + z b → x + z What should be the next step? Or is there a better method?

Question

Prove that angle bisectors of a triangle are concurrent using vectors. Also, find the position vector of the point of concurrency in terms of position vectors of the vertices.I solved this without using vectors to get some idea. I am not sure how to prove it using vectors. I don&#039;t want to use vector equations for straight lines and then find the point of concurrency. That&#039;s like solving using coordinate geometry.Let             a      →        ,            b      →        ,            c      →       represent the sides   A  ,  B  ,  C respectively.The angle bisectors are along                               a          →                                      |                                      a            →                                    |                      +                              b          →                                      |                                      b            →                                    |                      ,                                b          →                                      |                                      b            →                                    |                      +                              c          →                                      |                                      c            →                                    |                      ,                                c          →                                      |                                      c            →                                    |                      +                              a          →                                      |                                      a            →                                    |                    Let the sides   A  B  ,  B  C  ,  C  A be   x  ,  y  ,  z. Let   A  D be one of the angular bisector.            B      D              C      D        =      x    z  Hence  D  =            x                        c          →                    +      z                        b          →                            x      +      z      What should be the next step? Or is there a better method?

Tianna Deleon · Accepted Answer

You just need to apply the Angle Bisector Theorem again. For completeness, I&#039;ll go through the whole argument.Given   △  A  B  C with side-lengths   a  =      |              B      C        →        |  ,   b  =      |              C      A        →        |  ,   c  :=      |              A      B        →        |  , suppose the angle bisector from   A meets the opposite side at   D. You&#039;re correct that, by the Angle Bisector Theorem                              (                      ⋆                    )                                                                |                                                      B                D                            →                                      |                                                          |                                                      D                C                            →                                      |                                      =                                            |                                                      A                B                            →                                      |                                                          |                                                      A                C                            →                                      |                                      =                  c          b                    so that we can write:  D  =            B      b      +      C      c              b      +      c      Note also that   (  ⋆  ) implies                    |                              B          D                →                    |                            |                              B          C                →                    |              =      c          b      +      c          so that        |              B      D        →        |    =      |              B      C        →        |        c          b      +      c        =            a      c              b      +      c      Now, here&#039;s your next step: Suppose the angle bisector from   B meets             A      D        ¯   at   X (which we &quot;know&quot; is the incenter). Again by the Angle Bisector Theorem applied to   △  A  B  D,                    |                              A          X                →                    |                            |                              X          D                →                    |              =                    |                              B          A                →                    |                            |                              B          D                →                    |              =      c          a      c              /            (      b      +      c      )        =            b      +      c        a  and we can write  X  =            A      a      +      D      (      b      +      c      )              a      +      b      +      c        =            A      a      +      B      b      +      C      c              a      +      b      +      c      Because this formula is symmetric in the elements of   △  A  B  C, we conclude that   X is not merely where the angle bisector from   A meets the angle bisector from   B, but where any two angle bisectors meet ... and, therefore, where all three bisectors meet (aka, the incenter).

Prove that angle bisectors of a triangle are concurrent using vectors. Also, find the position vecto

Answered question

Answer & Explanation

New Questions in Elementary geometry