Let the vertexes be A, B, C respectively.
1) Choose a vertex of the triangle (e.g. A), and construct the (interior) angle bisector of the angle CAB at that vertex.
2) Mark the intersection of that angle bisector and the opposite edge BC as O. This shall be your circle's centre.
3) Construct lines perpendicular to AB and AC and passing through O, let the intersections be D and E.
We now have a circle that is centred at O and passes through D and E. Furthermore this circle is tangent to AB and AC at D and E.
This method works because we use the fact that angles CAO and OAB have to be equal if the circle were to be tangent to AB and AC; this implies that the centre of the circle lies both on the angle bisector at A and the side BC, and thus centre of circle is well defined. Afterwards dropping perpendiculars from O onto AB and AC is using the property that a circle's tangent is perpendicular to the radius containing the point of tangency.