As shown in the figure, go through the center of gravity o of △ ABC (the intersection of the three midlines) and make a parallel line of BC, intersecting AB at D and AC at E. then the area ratio of △ ade to △ ABC is () A. 1：2B. 2：3C. 1：3D. 4：9

As shown in the figure, go through the center of gravity o of △ ABC (the intersection of the three midlines) and make a parallel line of BC, intersecting AB at D and AC at E. then the area ratio of △ ade to △ ABC is () A. 1：2B. 2：3C. 1：3D. 4：9

As shown in the figure, FH ∥ CG intersects AB at h through F, ∵ f is the midpoint of BC, ∥ BH = Hg = 12bg = 12ag, ∥ og ∥ FH, ∥ Ao: AF = Ag: ah = Ag: (Ag + Hg) = 1: (1 + 12) = 2:3, and ∵ de ∥ BC, ∥ ade ∥ ABC, Ao: AF = AE: AC, ∥ s △ ade: s △ ABC = AE2: ac2 = AO2: af2 = 4:9

In the triangle ABC, ab = AC = 3, BC = 2, the bisector of angle ABC intersects the parallel line of BC at D, and the area of triangle abd is calculated

3,3,2 triangles determined
High square 3 * 3-1 * 1 = 8
Area: 2 * root 8 divided by 2
It's two root two

In the triangle ABC, ab = AC = 3, BC = 2, the bisector of angle ABC intersects the parallel line of angle BC at point D, and the area of triangle ABC is calculated

3,3,2 triangles determined
High square 3 * 3-1 * 1 = 8
Area: 2 * root 8 divided by 2
It's two root two