As shown in the figure, in RT △ ABC, ∠ B = 90 °, ab = 1, BC = 12, take point C as the center, CB as the radius of arc intersection Ca at point D; take point a as the center, ad as the radius of arc intersection AB at point E. (1) find the length of AE; (2) take points a and E as the center, AB length as the radius of arc, two arcs intersect at point F (F and C are on both sides of AB), connect AF and EF, let the circle where EF intersects de at point G, connect AG, try to guess Think about the size of EAG and explain the reason

As shown in the figure, in RT △ ABC, ∠ B = 90 °, ab = 1, BC = 12, take point C as the center, CB as the radius of arc intersection Ca at point D; take point a as the center, ad as the radius of arc intersection AB at point E. (1) find the length of AE; (2) take points a and E as the center, AB length as the radius of arc, two arcs intersect at point F (F and C are on both sides of AB), connect AF and EF, let the circle where EF intersects de at point G, connect AG, try to guess Think about the size of EAG and explain the reason

(1) In RT △ ABC, from ab = 1, BC = 12, AC = 12 + (12) 2 = 52, ∵ the arc intersection CA with point C as the center and CB as the radius is at point D; the arc intersection AB with point a as the center and ad as the radius is at point e ∵ BC = CD, AE = ad, ∵ AE = ac-cd = 5 − 12; (2) ∠ EAG = 36 ° for the following reasons: ∵ FA = Fe = AB = 1, ae