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Detailed explanation of 25 questions in the senior high school entrance examination of Shanghai mathematics in 2009 ~ ~
Solution: (1)AD=2, and point Q coincides with point B. According to the meaning of the question, ∠PBC=∠PDA because ∠A=90. PQ/PC=AD/AB= 1, so: △PQC is an isosceles right triangle, BC=3, so: PC=3/2,

(2) Add auxiliary lines as shown in the figure. According to the meaning of the question, the areas of two triangles can be expressed as S 1, S2,? The height is h, h,

Then: s1= (2-x) h/2 = (2 * 3/2)/2-(x * h/2)-(3/2) * (2-h)/2.

S2=3*h/2 because of two S 1/S2=y, eliminate h, h, and get:

Y=-( 1/4)*x+( 1/2),?

Domain: When point P moves to coincide with point D, the value of X is the largest. When PC is perpendicular to BD, then X=0. Connect DC to make QD a vertical DC. According to the known conditions, the circle with four points B, Q, D and C can be found. It can be inferred from the theorem of circle angle that the triangle QDC is similar to the triangle ABD.

QD/DC=AD/AB=3/4, let QD=3t and DC=4t, then: QC=5t, which is obtained by Pythagorean theorem:

In the right triangle AQD: (3/2) 2+(2-x) 2 = (3t) 2.

QBC: 3 2+x 2 = (5t) 2 in a right triangle.

Finishing: 64x 2-400x+30 1 = 0? (8x-7)(8x-43)=0

X 1=7/8? X2=(43/8)>2 (give up)? So function:

The domain of Y=-( 1/4)*x+ 1/2 is [0,7/8].

(3) Because PQ/PC=AD/AB, assuming PQ is not perpendicular to PC, we can draw a straight line PQ' perpendicular to PC and intersecting with AB at Q'.

Then, four-point * * circles of B, Q', P and C are obtained from the theorem of the angle of circle and the properties of similar triangles;

PQ′/PC = AD/AB,

Since PQ/PC=AD/AB and point Q' coincide with point Q, the angle ∠ QPC = 90.