In Δ A B C , A C &gt; A B .. The internal angle bisector of ∠ A meets BC at D, and E is the foot of the perpendicular from B onto AD. Suppose A B = 5 , B E = 4 and A E = 3. Find ( A C + A B A C − A B ) E D.

Question

In   Δ  A  B  C  ,  A  C  &amp;gt;  A  B  .. The internal angle bisector of   ∠  A meets BC at D, and E is the foot of the perpendicular from B onto AD. Suppose   A  B  =  5  ,  B  E  =  4 and   A  E  =  3. Find             (                  A      C      +      A      B              A      C      −      A      B                  )        E  D.

nuramaaji2000fh · Accepted Answer

Step 1I use a bit of trigonometry.(I would love to see something without this)Let   B  C  =  a  ,  C  A  =  b  ,  A  B  =  c.By Napiers Analogy:                    (1)                    tan        ⁡                                            B              −              C                        2                          =                              b            −            c                                b            +            c                          cot        ⁡        (        A                  /                2        )            Notice that  sin  ⁡  (  A      /    2  )  =      4    5                      (2)                    cot        ⁡        (        A                  /                2        )        =        ?            Step 2Also   ∠  E  D  B  =  90  +            C      −      B        2                      (3)                    tan        ⁡                  ∠          E          D          B                =        tan        ⁡        (        90        +                              C            −            B                    2                )        =        cot        ⁡                                            B              −              C                        2                          =                  4                      E            D

Jaxon Hamilton · Accepted Answer

Step 1This proof can probably be simplified or made more natural, but here is my take:Construct CG parallel to FB as shown above. We have:  △  A  F  E  ≅  △  A  B  E     (  A  S  A  )  ,    △  A  F  E  ∼  △  A  C  G  ,  △  B  E  D  ∼  △  C  G  D     (  A  A  A  )The last ratio is by Angle Bisector Theorem.Step 2Hence:             A      C      +      A      B              A      C      −      A      B        =            A      C              C      F        +            A      B              C      F        =            A      B              C      F            (    1    +                  A        C                    A        B              )    =            A      F              C      F            (    1    +                  D        G                    E        D              )    =            A      E              E      G                  E      G              E      D        =            A      E              E      D      Giving the result:       (                  A        C        +        A        B                    A        C        −        A        B              )    E  D  =  A  E

In triangle ABC, AC>AB. The internal angle bisector of angle A meets BC at D, and E is the foot of the perpendicular from B onto AD. Suppose AB=5, BE=4 and AE=3 . Find ((AC+AB)/(AC−AB))ED .

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