(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.