Find angle of isosceles triangleLet ABC an isosceles triangle ( A B = A C) and the bisector from B intersects AC at D such that A D + B D = B C. Find the angle A.

Question

Find angle of isosceles triangleLet ABC an isosceles triangle (  A  B  =  A  C) and the bisector from B intersects AC at D such that   A  D  +  B  D  =  B  C. Find the angle A.

plomet6a · Accepted Answer

Step 1Let E be placed on BC such that   E  C  =  A  D.Thus,   B  D  =  B  C  −  A  D  =  B  C  −  E  C  =  B  E  ,, which says   ∡  B  E  C  =      1    2    (      180          ∘        −  ∡  D  B  C  )  =      1    2    (      180          ∘        −      45          ∘        +  α  )  =      67.5          ∘        +      1    2    α  ..Step 2In another hand, since BD is a bisector of   Δ  A  B  C, we obtain:            E      C              A      B        =            A      D              A      B        =            D      C              B      C       and   ∠  C is a common angle of   Δ  A  B  C and   Δ  E  D  C  ..Thus,   Δ  A  B  C  ∼  Δ  E  D  C  ,, which gives   ∡  D  E  C  =  4  α.Id est,       67.5          ∘        +      1    2    α  +  4  α  =      180          ∘        ,, which gives   α  =      25          ∘       and   ∡  B  A  C  =  4  ⋅      25          ∘        =      100          ∘        ..

propappeale00 · Accepted Answer

Step 1First, let&#039;s see the angles-   ∡  A  B  D  =                              180          ∘                −        4        a            4        =      45    ∘    −  a.-   ∡  A  D  B  =      180    ∘    −  4  a  −  (      45    ∘    −  a  )  =      135    ∘    −  3  a.-   ∡  B  D  C  =      180    ∘    −  (      135    ∘    −  3  a  )  =      45    ∘    +  3  a-   ∡  C  =      90    ∘    −  2  aFrom the theorem of sines in   Δ  A  B  D:            A      D      +      B      D              A      B        =            sin      ⁡      (      ∡      A      B      D      )      +      sin      ⁡      (      ∡      A      )              sin      ⁡      (      ∡      A      D      B      )        =            sin      ⁡      (              45        ∘            −      a      )      +      sin      ⁡      4      a              sin      ⁡      (              45        ∘            +      3      a      )      and from the theorem of sines in   Δ  A  B  C:            B      C              A      B        =            sin      ⁡      (      ∡      A      )              sin      ⁡      (      ∡      C      )        =            sin      ⁡      4      a              sin      ⁡      (              90        ∘            −      2      a      )        =            sin      ⁡      4      a              cos      ⁡      2      a        =            2      sin      ⁡      2      a      cos      ⁡      2      a              cos      ⁡      2      a        =  2  sin  ⁡  2  abecause the cos function is dephased from sin by       90    ∘  . Now from the hypothesis:            sin      ⁡      (              45        ∘            −      a      )      +      sin      ⁡      4      a              sin      ⁡      (              45        ∘            +      3      a      )        =  2  sin  ⁡  2  aand this implies   sin  ⁡  4  a  +  sin  ⁡      (          45              ∘              −    a    )    =  2  sin  ⁡  2  a  sin  ⁡      (          45              ∘              +    3    a    )   or   sin  ⁡  4  a  +  cos  ⁡      (          45              ∘              +    a    )    =  cos  ⁡      (          45              ∘              +    a    )    −  cos  ⁡      (          45              ∘              +    5    a    )   or   sin  ⁡  4  a  =  sin  ⁡      (    5    a    −          45              ∘              )   and thus   4  a  +  5  a  −      45          ∘        =      180    ∘    ⇒  a  =      25    ∘  . Which means   ∡  A  =      100    ∘

Let ABC an isosceles triangle (AB=AC) and the bisector from B intersects AC at D such that AD+BD=BC. Find the angle A.

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