第三个问，为什么当AP//BD时，F就是PG的中点？怎么证明呢？谢谢

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Demonstration:

Let's build the center circle F middle of PG and going through G: either: c-Circle(F,G)

Let's write the function in several variables:

CE(x,y)=(x - x(F))² + (y - y(F))² - Radius(c)²

Points D, G and P check relationships:

CE(x(D),y(D)) = CE(x(G),y(G)) = CE(x(P),y(P)) = 0

They are therefore cocyclical on circle c.

And this regardless of the position of the point G. 1 1

F 点并不总是 FG 的中间点！

F 位于 FG 中间的唯一条件是 P、G 和 E 点位于相同的 PG 直径圆（这些点是共周期的）。 1

Thank you very much for your patience. Ask you one more question, below.

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The parabola y = a (x + 2) ^ 2 + c intersects the X axis at A and B, and intersects the Y axis at C. We know that point A (-1,0), OB = OC.

(1) Write the analytical expression of the parabola

(2) Point Q is a point on the straight line y = -x-4. Is there a point P on the parabola's axis of symmetry such that ∠APB = 2∠AQB and there is only one such point Q? If present, the coordinates of the point P are requested. If not, please explain why.

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Correct the title

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(2) Point Q is a point on the straight line y = -x-4. Is there a point P on the parabola's axis of symmetry such that ∠APB = 2∠AQB and there is only one such point Q? If it exists, find the coordinates of point P. If not, please explain why. 1

There is indeed a P point on the parabola's axis of symmetry, and even an infinity by moving the Q point on the right y-x-4,

which verifies the relationship 'APB' to 2'AQB.

Let's trace the mediator of the AQ segment, which cuts the axis of symmetry to a P point.

Let's build the P center circle through the Q, A and B points, the APB and AQB angles

check the relationship with ABP and AQB angles (because ABP is the angle in the centre and AQB underlies the same AB arc).

The coordinates of point P are that of the center of the circle. 