① This problem can be solved by constructing trigonometric function and equilateral triangle, using the properties of right triangle and equilateral triangle, congruent triangles's judgment and properties, and similar triangles's judgment and properties;

② Both △ ABC and △CDE are general oblique triangles, so it is not easy to get the results directly according to the known conditions. However, since AC in △ ABC is known and ∠BAC = 60, if a right triangle or an equilateral triangle is formed with AC as one side and ∠BAC as-inner angle, the areas of two triangles can be obtained.

explain

Solution 1:

As shown in the figure: the vertical line intersecting C and AB intersects the extension line of AB at G.

E is the midpoint of BC.

∴BE=CE=GE

∴∠GBC=∠BGE=80

∠∠ABC = 100，∠DEC=80，∠A=60

∴∠BCA=20，∠EDC=80

∴△CDE≌△EBG

∴S△BGE=S△DEC

E is the midpoint of BC.

∴S△BGC=2S△BGE

∴2S△CDE=S△CBG

∴S△ABC+2S△CDE

=S△ABC+S△CBG

=S△CGA

=( 1/2)AG？ Democratic Republic of the Congo

=√3/8

This is the solution to form a right triangle.

Solution 2:

As shown in the figure, with AC as one side and ∠BAC as the-inner angle, a positive △ACG is formed.

The bisector of ∠GCB passes through GA in f,

Then S△GAC

=( 1/2)AC2？ sin60

=√3/4

Prove △ BAC △ FGC, △CED∽△CBF.

∫CE =( 1/2)BC

∴S△CED=( 1/4)S△CFB

∴S△ABC+2S△CDE

=S△ABC+( 1/2)S△CFB

=( 1/2)S△CGA

=√3/8