As shown in the figure, in the isosceles △ ABC, AC = BC = 10, take BC as the diameter, make ⊙ o, intersect AB at point D, intersect AC at point G, DF ⊥ AC at point F, and intersect the extension line of CB at point E. (1) prove that the straight line EF is the tangent of ⊙ o; (2) if sin ∠ e = 25, find the length of ab

As shown in the figure, in the isosceles △ ABC, AC = BC = 10, take BC as the diameter, make ⊙ o, intersect AB at point D, intersect AC at point G, DF ⊥ AC at point F, and intersect the extension line of CB at point E. (1) prove that the straight line EF is the tangent of ⊙ o; (2) if sin ∠ e = 25, find the length of ab

(1) It is proved that: connecting OD, ∵ AC = BC, ∵ ABC = ∵ BAC, ∵ od = ob, ∵ ABC = ∵ ODB, ∵ BAC = ∵ BDO, ∵ OD ∥ AC, ∵ DF ⊥ AC, ∵ OD ⊥ DF, ∵ od is the radius, and the ∵ straight line EF is the tangent of ⊙ o; (2) Connecting BG, ∵ BC is ⊙ o diameter, ∵ BGC = 90 °, ∵ DF ⊥ AC, ∵ DFC = 90 ° = ∵ BGC, ∵ BG ∥ EF, ∵ e = ∵ GBC, ∵ sin ∵ e = 25, ∵ sin ∵ GBC = 25 = CGBC, ∵ BC = 10, ∵ CG = 4, ∵ Ag = 10-4 = 6. According to Pythagorean theorem, BG = bc2-cg2 = 221. In RT △ BGA, ab = BG2 + ag2 = (221) 2 + 62 = 230, that is ab = 230

In the rectangular coordinate system, the coordinates of each vertex of △ OAB are a (- 3, - 4) B (5,0) o as the origin. (1) calculate the area of △ AOB. (2) calculate the distance from the origin to ab

(1)S△AOB=1/2*|OB|*|Ay|=1/2*5*4=10
(2) OA = 0b = 5, isosceles triangle, H = 1 / 2Ab, s △ AOB = AB * H = 2H * H = 10
H = root 5

In RT △ ABC, if the hypotenuse AB = 5 and the right angle BC = 5, then the area of △ ABC is______ ．

∵ in RT △ ABC, the hypotenuse AB = 5, the right angle BC = 5, the other straight angle AC is AB2 − BC2 = 20, and the area of ∵ ABC is 12 × AC × BC = 5

In RT △ ABC, if the hypotenuse AB = 5 and the right angle BC = 5, then the area of △ ABC is______ ．

∵ in RT △ ABC, the hypotenuse AB = 5, the right angle BC = 5, the other straight angle AC is AB2 − BC2 = 20, and the area of ∵ ABC is 12 × AC × BC = 5