Finding angle in a triangle with one median and an isosceles triangle in itIn the following figure A F = B D = D C and A E = E F. Find the angle α.

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Finding angle in a triangle with one median and an isosceles triangle in itIn the following figure   A  F  =  B  D  =  D  C and   A  E  =  E  F. Find the angle   α.

Mario Kerr · Accepted Answer

Step 1In isosceles   △  A  E  F, drop   E  G  ⊥  A  F with G on AF. Let   A  G  =  G  F  =  x. So   B  D  =  2  x. Let   ∠  B  A  D  =  θ. So   A  E  =  x  sec  ⁡  θ.Step 2Applying Menelaus&#039; for transversal CE to   △  A  B  D,             B      C              D      C        ⋅            D      F              F      A        ⋅            A      E              E      B        =  1  ⇒  D  F  =  E  B  cos  ⁡  θ  ⇒  A  D  −  2  x  =  (  A  B  −  x  sec  ⁡  θ  )  cos  ⁡  θ  ⇒  A  D  =  A  B  cos  ⁡  θ  +  xNext drop   B  H  ⊥  A  D with H on AD. In right ABH,   A  H  =  A  B  cos  ⁡  θ. Then   A  D  =  A  H  +  D  H implies   D  H  =  x!Hence in right triangle BHD,   B  D  =  2  x,   H  D  =  x  =  B  D      /    2. It turns out   △  B  H  D is       30          ∘        −      60          ∘        −      90          ∘       and we conclude   α  =      60          ∘

ignizeddyug · Accepted Answer

Step 1Draw the three lines a, b, c such that:  A    ∈    a       and       a        |        |      B  C  B    ∈    b       and       b        |        |      C  A  C    ∈    c       and       c        |        |      A  BFurthermore, let       B    ∗      =    a    ∩    c         and             A    ∗    =  b    ∩    c.Then, ABA∗C and ABCB∗ are parallelograms and furthermore   A  B  =  C      A    ∗    =  C      B    ∗   as well as   A      B    ∗    =  B  C.Step 2Apply Menelaus&#039; theorem to the triangle BCE intersected by the line that passes through the collinear points A,F,D:             A      B              A      E        ⋅            E      F              C      F        ⋅            D      C              B      D        =  1.Since by assumption   A  E  =  E  F and   B  D  =  C  D, one concludes that   A  B  =  C  F.Combining this latter fact with the earlier conclusions, one observes that   C  F  =  A  B  =  C      A    ∗    =  C      B    ∗   which is possible if and only if triangle   Δ        A    ∗        B    ∗    F has       90          ∘       angle at vertex F, i.e.   ∠        A    ∗    F      B    ∗    =      90          ∘      . But that means that:  ∠    A  F      B    ∗    =      180          ∘        −  ∠        A    ∗    F      B    ∗    =      180          ∘        −      90          ∘        =      90          ∘      Recall that by assumption   A  F  =  B  D  =  D  C  =      1    2      B  C. But by construction, ABCB∗ is a parallelogram and   A      B    ∗    =  B  C so   Δ    A  F      B    ∗       is a triangle with the properties that     ∠    A  F      B    ∗    =      90          ∘             and       A  F  =      1    2      A      B    ∗   which is possible if and only if   ∠    F  A      B    ∗    =      60          ∘      .However, AB∗ is parallel to BC and consequently, by the properties of a pair of parallel lines intersected by a third line,   ∠    A  D  B  =  ∠    D  A      B    ∗    =  ∠    F  A      B    ∗    =      60          ∘

In the following figure AF=BD=DC and AE=EF. Find the angle alpha.

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