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First: √6. Applying Pythagoras, it's easy to prove that AP²+CP²= BP²+DP².
Second: Assuming P is within ABCD (if it's not, see below) rotate the square 90°, with center B, so that A moves to A', D moves to D', C moves to A, and P moves to P'. Angle PBP'=90° by definition. Since PB=P'B, triangle PBP' is isosceles, and thus angle P'PB=45°. From Pythagoras, PP'=√8. As AP'=CP=3, it follows that APP' is a rectangular triangle, and angle PP'A=90°. Summing both results, angle APB=135°. If P lies outside the square, the same reasoning would apply, but then APB would be 90°-45°=45°. |
