Question of Triangle of GeometryIn triangle ABC,we have A B &gt; A C. If A' is the mid point of BC, AD is the altitude through A and if the internal and external bisectors of angle A meet at BC at X and X' respectively ,prove that A ′ D = ( c 2 − b 2 ) / 2 aI have tried to use obtuse angle theorem to find height then solve the question but stuck on double variable.

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Question of Triangle of GeometryIn triangle ABC,we have   A  B  &amp;gt;  A  C. If A&#039; is the mid point of BC, AD is the altitude through A and if the internal and external bisectors of angle A meet at BC at X and X&#039; respectively ,prove that       A    ′    D  =  (      c    2    −      b    2    )      /    2  aI have tried to use obtuse angle theorem to find height then solve the question but stuck on double variable.

bulgarum87 · Accepted Answer

Step 1Since   c  &amp;gt;  b, we obtain   γ  &amp;gt;  β  ,, which says that   ∠  A  B  C is an acute angle.Thus,   B  D  =  c  cos  ⁡  β  =            c      (              a        2            +              c        2            −              b        2            )              2      a      c        =                    a        2            +              c        2            −              b        2                    2      a        .Step 2Also, me see that   B  D  &amp;gt;  B      A    ′   because   c  &amp;gt;  b.Thus,       A    ′    D  =  B  D  −  B      A    ′    =                    a        2            +              c        2            −              b        2                    2      a        −      a    2    =                    c        2            −              b        2                    2      a        .

In triangle ABC,we have AB>AC. If A' is the mid point of BC, AD is the altitude through A and if the internal and external bisectors of angle A meet at BC at X and X' respectively, prove that A′D=(c^2-b^2)/(2a).

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