Mathematics self-test paper answer

(Beijing simulation) In △ABC, a semicircle O 1 and a semicircle O2 are made outside △ABC with AB and AC as diameters respectively, where O 1 and O2 are the centers of two semicircles respectively. F is the midpoint of side BC, and points D and E are the midpoints of two semi-circular arcs respectively.

(1) proof: as shown in figure 1, connect O 1F, O 1D, DF, O2F, O2E, EF, proof: △ do1f △ fo2e;

(2) As shown in Figure 2, intersection point A is tangent to semicircle O 1 and semicircle O2, respectively, and the extension lines of BD and CE intersect at point P and point Q to connect PQ. If ∠ ACB = 90, DB = 5 and CE = 3, find the length of the straight line PQ;

(3) As shown in Figure 3, intersection A is the tangent of semicircle O2, the extension line of intersection CE is at point Q, intersection Q is the perpendicular of straight line FA, and the extension line of intersection BD is at point P, connecting PA. It is proved that PA is the tangent of a semicircle O 1.

(1)∫o 1, O2 and F are the midpoint of AB, AC and BC respectively.

∴O 1F∥AC and O 1F=AO2, o2f∨ab and O2F=AO 1,

∴∠BO 1F=∠BAC,∠CO2F=∠BAC，

∴∠BO 1F=∠CO2F

Point d and point e are the midpoint of two semi-circular arcs,

∴o 1f=ao2=o2e,o2f=ao 1=o 1d，

∠BO 1D=90，∠CO2E=90，

∴∠BO 1D=∠CO2E.

∴∠DO 1F=∠FO2E.

∴△do 1f≌△fo2e；

(2) Solution: As shown in Figure 2, extend CA to G, make AG=AQ, and connect BG and AE.

Point e is the midpoint of the semicircular O2 arc,