Question 1123946
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    < ADB = 180° - (half of < A + half of < B)     (1)    (as the sum of interior angles of the triangle ABD)



From the other side

      < A + < B + < C = 180°  ====>  half of < A + half of < B = 90° - half < C.



Therefore, you can continue (1) in this way :


    < ADB = 180° - (half of < A + half of < B) = 180° - (90° - half < C) = 90° + half < C.



Or, in terms of  {{{alpha}}},  {{{beta}}}  and  {{{gamma}}}


    < ADB = {{{180^o - 0.5*(alpha + beta)}}} = {{{90^o + 0.5*gamma}}}.
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Solved.