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(2)P0、Q0、Q1、P1四点共圆.(1)证明略.图1(2)证法1:如图1,联结P0P1、P0Q0、P0Q1、P1Q0、P1Q1、Q0Q1.由题设可知,△P0B0Q0、△P1B1Q1、△P0C0Q1、△P1C1Q0均为等腰三角形.设∠P0B0Q0=∠AB0C1=α,∠P1C1Q0=β.易知∠P0Q0P1=∠P0Q0B0+∠P1Q0C1=2π-2α+2π-2β=π-21(α+β)=2π+
(2) P0, Q0, Q1, P1 four points of a total of the circle. (1) prove slightly. It can be seen that △ P0B0Q0, △ P1B1Q1, △ P0C0Q1, △ P1C1Q0 are isosceles triangle. Let ∠P0B0Q0 = ∠AB0C1 = α, ∠P1C1Q0 = β. It is easy to know ∠P0Q0P1 = ∠P0Q0B0 + ∠P1Q0C1 = 2π-2α + 2π-2β = π-21 (α + β) = 2π +