To prove that AD is tangent to the circle passing through H with center the midpoint of BC, we will use the fact that if a line is tangent to a circle, then it is perpendicular to the radius at the point of tangency.

Let O be the midpoint of BC, and let E be the intersection of AH and the circle with center O. Since O is the midpoint of BC, DE = DH (because they are both radii of the same circle). Also, since H is the orthocenter, ∠BHA = 90° - ∠C.

Now consider triangle AHD. We have:

∠AHD = 180° - ∠AHB - ∠DHA
= 180° - (90° - ∠C) - ∠C
= 90°

Since ∠AHD = 90°, AD ⊥ HD. But HD is a radius of our circle at point D. Therefore, AD is tangent to the circle passing through H with center O (the midpoint of BC).
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