Prove PM is Diameter in ΔABC with ∠A = 90°

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In triangle ABC, with angle A as 90 degrees, M is the midpoint of side BC and H is the foot of the altitude from A to BC. A circle l is drawn through points A, M, and C, and a line through M perpendicular to AC intersects AC at D and circle l again at P. The key to proving PM is a diameter lies in recognizing that D, as the midpoint of chord AC, ensures that PM must be a diameter of circle l. The similarity of triangles MCD and MPC, along with the relationships established by angles and midpoints, supports this conclusion. Thus, the proof that PM is the diameter follows directly from the properties of the circle and the right triangle configuration.
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Homework Statement


In ΔABC, ∠A = 90°, M is the midpoint of BC and H is the foot of the altitude from A to BC. A circle l is drawn through points A, M and C. The line passing through M perpendicular to AC meets AC at D and the circle l again at P. BP intersects AH at K.

Prove that PM is diameter.

The Attempt at a Solution


No idea how to do at all. Any help would be appreciated, as this is part 1 of a long question.
 

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okay, but what are you trying to figure out?
 
Opps sorry. I want to figure out that PM is the diameter
 
Have found solution but need further help! Here is my attempt:

Found PM is diameter of circle l, Triangle MCD similar to Triangle MPC, Triangle DMB similar to Triangle BMP, Angle DBM = Angle ABK.

Prove AK = KH, (using similar triangles or otherwise)
 
There seems to be a lot of extraneous information here. To prove that PM is a diameter of circle l[/tex], all you need to do is note that you are given a right triangle (\triangle ABC[/tex]) with M as the midpoint of BC. Therefore, D is the midpoint of AC (which can be shown since \triangle ABC[/tex] is similar to \triangle DMC[/tex]). AC is a chord to circle l[/tex] and since D is the midpoint of that chord, PM has to be a diameter.
 

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