• anonymous
From a point P, 2 tangents, PA and PB are drawn to a circle with centre "O". If OP is equal to diameter of a circle, prove that triangle PAB is an equilateral traingle.
  • Stacey Warren - Expert
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  • schrodinger
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  • anonymous
|dw:1328520948852:dw| look i relplaced P by c and p is the center of circle The distances from the point of intersection of triangle bisectors to all three triangle sides are equal. Therefore, this point is the incircle center as well. Now, let F, G, and H are the midpoints of AB, BC, and AC, respectively. And let P be the point where perpendiculars going through F and G cross. Triangles AFP and FBP are equal (S-A-S), so are triangles BGP and GCP (Side-Angle-Side). Therefore, AP = BP and BP = CP. PH is perpendicular to AC because triangles APH and CPH are equal (Side-Side-Side) and angles AHP and CHP are bothe equal and complimentary.

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