Point O is marked in triangle ABC at height BD; ∟ОАD = ∟ОСD. Prove that point O is equidistant from lines AB and BC

Since, according to the condition, the angle OAD = OСD, then the AOС triangle is isosceles, OA = OС.
Then the segment OD is its height, bisector and median, which means the angle AOD = СOD.
Then the angle AOB = COB as adjacent angles equal to the angle.
In triangles AOB and COB AO = CO, side OB is common, angle AOB = COB, then triangles AOB and COB are equal on two sides and the angle between them.
The heights of equal triangles drawn to similar sides are equal, then OK = OM, as required.