To prove that AD is tangent to the circle passing through H with center the midpoint of BC, we need to show that the angle between AD and the tangent line is 90 degrees. Let's proceed with the proof.

Let M be the midpoint of BC.
Consider the circle passing through H with center M. Let's denote this circle as O.
Since O passes through H and has its center at the midpoint of BC, we know that OH is a diameter of the circle.
It follows that angle OMH is a right angle.
We need to show that angle HAD is also a right angle to prove that AD is tangent to the circle O.
Since ABC is an acute triangle, we know that the altitude AD is inside the triangle.
Let E be the intersection of the line AH with the circle O. We need to show that E lies on AD.
Since AH is a line passing through A and E, and AD is perpendicular to BC, we have angle HAD = 90 degrees if and only if angle EAD = 90 degrees.
To show that angle EAD = 90 degrees, it suffices to prove that AE is perpendicular to AD.
Consider the triangle AHE. We have angle HAE = 90 degrees because AH is the altitude of triangle ABC.
Since OH is a diameter of circle O, angle OMH = 90 degrees.
Therefore, angles HAE and OMH are equal.
It follows that triangle AHE is similar to triangle OMH.
Since corresponding angles in similar triangles are equal, we have angle AHE = angle OMH.
But angle OMH is a right angle, so angle AHE is also a right angle.
Therefore, AE is perpendicular to AD, which implies that angle EAD = 90 degrees.
This shows that AD is tangent to the circle passing through H with center the midpoint of BC.
Hence, we have proven that AD is tangent to the circle passing through H with center the midpoint of BC.
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