(1) Find the radius of ⊙O;
(2) Verification: AE = BF
(3) When AE= 1, is there a point P on the line segment AB? Is the triangle of points A, P and D similar to the triangle of points B, P and C? If it exists, track the positions of all points p in the graph that meet the conditions (no calculation is needed); If it does not exist, please give reasons.
(4) When AE is what value, only two points P can satisfy the condition in (3)?
Test site: right-angled trapezoid; Vertical diameter theorem; Similar triangles's judgment and nature.
Analysis: (1) connecting DG, DG⊥BC can be obtained according to the right angle of the circumference of the diameter, and the length of DC can be obtained according to the pythagorean theorem in △DGC;
(2) Let OM⊥AB be m, calculate EM=FM according to the vertical diameter theorem, and deduce am = bm according to the trapezoid midline.
(3) There are three points: ① When P and E coincide ∠ CED = 90, ∠AED=∠ECB and ∠DAB=∠ABC are obtained according to the equivalence of the complementary angles of the same angle, and △AED∠△BCE is proved by the similarity of two triangles with equal corresponding angles. (2) When point P and point F coincide, similar to (1), AP can be found; ③P On the line EF, we can get AP from △APD∽△BPC according to the properties of similar triangles.
(4) When P3 coincides with E(P 1), that is, ∠ AED = ∠ BEC = 45, there are only two solutions, and AE = 3 is obtained by the proportional formula according to the properties of similar triangles.
Solution: Solution: (1) Connect DG.
∫CD is the diameter,
∴DG⊥BC，
At △DGC, ∫BC-AD = 1,
∴GC= 1，
AB = 7，
∴DC= 72+ 12=5 2，
∴⊙O has a radius of 522;
(2) Let OM⊥AB be m, and get EM=FM according to the vertical diameter theorem.
And ∵AD∨OM∨BC, OD=DC.
∴AM=BM，
∴AM-EM=BM-FM，
That is AE = BF.
(3) There are three points.
Let AD=x and BC=x+ 1. According to Pythagorean theorem,
AD2+AE2=DE2, namely x2+ 1=DE2,
BE2+BC2=CE2, which means 62+(x+ 1)2=CE2.
CE2+DE2=CD2=50，
That is x2+1+[62+(x+1) 2] = 50,
The solution is x=2,
That is, AD=2 and BC = 3.
Situation 1: ∠ APD+∠ BPC = 90.
Only ∠DPC=90 degrees, ∠APD+∠BPC=90, △ pad ∽△ CBP.
According to the characteristics of the circle, CD is the diameter, so such points are all on the arc, that is, point E and point F.
Let AF = Y. Then according to AD2+AF2+BF2+BC2=CD2.
∴4+y2+(7-y)2+9=50，
The solution is y 1= 1, and y2=6.
That is, AP= 1 or AP = 6；;
In the second case: ∠APD=∠BPC, the triangular pad is similar to PBC. ..
Suppose there is such a point P:∠APD =∠BPC, △APD∽△BPC, then
AP:BP=AD:BC=2:3，
∫AP+BP = AB = 7，
So AP = AP=7× 25= 145. ..
To sum up, it can be seen that there are three such points, and the lengths of AP are 1, 6,145 respectively;
(4) When P3 coincides with E(P 1), that is, ∠ AED = ∠ BEC = 45, then △APD and △BPC are isosceles right triangles.
From △APD∽△BPC, AP=AD and BP=BC,
AP+BP=7，BC-AD= 1，
AP = 3, which means AE = 3.
Therefore, when AE=3, only two points P satisfy the condition in (3), namely point E and point F. 。