Proof of angle Bisector Theorem by VectorsFor Δ A B C, let AD is its internal angle bisector, then we know that A B A C = B D C D and I know that it can be proved by geometry by drawing a line through C parallel to AD and extend AB so that it cuts new line and then use various angles and concept of similar triangle to prove our result. But I was wondering if we can prove it with the help of vectors as well?If we take position vectors of A,B and C as 0 → , b → and c → respectively , then A D → can be taken as λ ( b ^ + c ^ ). Could someone help me to proceed after that or provide some alternate approach through vectors?

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Proof of angle Bisector Theorem by VectorsFor   Δ  A  B  C, let AD is its internal angle bisector, then we know that             A      B              A      C        =            B      D              C      D       and I know that it can be proved by geometry by drawing a line through C parallel to AD and extend AB so that it cuts new line and then use various angles and concept of similar triangle to prove our result. But I was wondering if we can prove it with the help of vectors as well?If we take position vectors of A,B and C as       0    →    ,      b    →   and       c    →   respectively , then             A      D        →   can be taken as   λ  (            b      ^        +            c      ^        ). Could someone help me to proceed after that or provide some alternate approach through vectors?

neobuzdanio · Accepted Answer

Step 1As mentioned,                     A        D            →        ≠  λ  (            b      ^        +            c      ^        ) in general. I think there&#039;s still a decent looking proof using dot product.As usual, it helps if you sketch a diagram.Let&#039;s suppose   ∠  B  A  D  =  ∠  C  A  D  =  θ, where   θ is acute, and   |                    B        D            →        |  :  |                    C        D            →        |  =  r  :  s. Using the section formula,                     A        D            →        =      r          r      +      s                          A        C            →        +      s          r      +      s                          A        B            →        .Then by dot product,                     A        D            →        ⋅                    A        C            →        =  |                    A        D            →        |  |                    A        C            →        |  cos  ⁡  θ  =      r          r      +      s        |                    A        C            →            |    2    +      s          r      +      s                          A        B            →        ⋅                    A        C            →                          (1)                            ⟹                |                                            A              D                        →                          |        cos        ⁡        θ        =                  r                      r            +            s                          |                                            A              C                        →                          |        +                  s                      r            +            s                          |                                            A              B                        →                          |        cos        ⁡        2        θ        .            Step 2Similarly, by considering                     A        D            →        ⋅                    A        B            →      , we have                     (2)                    |                                            A              D                        →                          |        cos        ⁡        θ        =                  r                      r            +            s                          |                                            A              C                        →                          |        cos        ⁡        2        θ        +                  s                      r            +            s                          |                                            A              B                        →                          |        .            Combining 1 and 2 gives                               r                      r            +            s                          |                                            A              C                        →                          |        +                  s                      r            +            s                          |                                            A              B                        →                          |        cos        ⁡        2        θ                            =                  r                      r            +            s                          |                                            A              C                        →                          |        cos        ⁡        2        θ        +                  s                      r            +            s                          |                                            A              B                        →                          |                                      r                      r            +            s                          |                                            A              C                        →                          |        (        1        −        cos        ⁡        2        θ        )                            =                  s                      r            +            s                          |                                            A              B                        →                          |        (        1        −        cos        ⁡        2        θ        )                                                  |                                                            A                  B                                →                                      |                                |                                                            A                  C                                →                                      |                                              =                  r          s                                                  =                              |                                                            B                  D                                →                                      |                                |                                                            C                  D                                →                                      |                          .

For triangle ABC, let AD is its internal angle bisector, then we know that (AB)/(AC)=(BD)/(CD)

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