In ∆ABD, ∠BAD = (∠BAC)/2 and ∠ABD = (∠ABC)/2.
The sum of angles in that triangle is 180°, so
... ∠ADB + ∠BAD + ∠ABD = 180°
Substituting the above relations, we have ...
... ∠ADB + (∠BAC)/2 + (∠ABC)/2 = 180°
... ∠ADB = 180° - (∠BAC + ∠ABC)/2
Turning our attention now to ∆ABC, its angles, too, add to 180°.
... ∠BAC + ∠ABC + γ = 180°
We can subtract γ to get
... ∠BAC + ∠ABC = 180° - γ
Substituting this expression into the above expression for ∠ADB, we have
... ∠ADB = 180° - (180° - γ)/2
... ∠ADB = 90° + γ/2
