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Circle Geometry

  • Thread starter Lukybear
  • Start date
  • #1
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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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Answers and Replies

  • #2
277
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okay, but what are you trying to figure out?
 
  • #3
8
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Opps sorry. I want to figure out that PM is the diameter
 
  • #4
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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)
 
  • #5
754
1
There seems to be a lot of extraneous information here. To prove that PM is a diameter of circle [itex]l[/tex], all you need to do is note that you are given a right triangle ([itex]\triangle ABC[/tex]) with M as the midpoint of BC. Therefore, D is the midpoint of AC (which can be shown since [itex]\triangle ABC[/tex] is similar to [itex]\triangle DMC[/tex]). AC is a chord to circle [itex]l[/tex] and since D is the midpoint of that chord, PM has to be a diameter.
 

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