Shanghai Mathematics 200 1 Solution to the Last Question in the Final Exam

(1)∵∠BPC=∠A ∠APB=∠PBC (two straight lines are parallel with equal internal angles) ∴∠ABP=∠PCB.

∠∠PCB =∠DPC (two parallel lines with equal internal dislocation angles) ∴∠ABP=∠DPC (equivalent substitution)

∠∠a=∠∴△ Headquarters base ∠△DPC∴ab/DP = AP/DC.

Let AP=x, then DP=5-x, and we can get an equation about X as follows:

2/(5-x)=x/2 gives x 1= 1 x2=4.

So the value of AP is 1 or 4.

(2)① It is easy to get △APB∽△CEQ.

So AB/CE=AP/CQ ① and CE/DP=QC/QD (this can be deduced from CE=CQ? DP/QD)

Substitute into the formula 1 because ab = 2ap = xcq = ydp = 5-xqd = 2+y.

Y=- 1/2x2+5/2x-2 ② After finishing.

② AB=AP/CQ y=x/2③ Because CE= 1.

② ③ The formula "X2-4x+4" is obtained by sorting at the same time, so x=2.

So the AP length is 2 at this time.

I hope to accept (the square of x behind, here I hope to see clearly that x2 refers to the square of x)