In △ ABC, ∠ ACB = 90 ° ch ⊥ AB in H, △ ACD and △ BCE are equilateral triangles. (1) prove: △ Dah ∽ ECH; (2) if ah: HB = 1:4, find s △ Dah: s △ ECH

In △ ABC, ∠ ACB = 90 ° ch ⊥ AB in H, △ ACD and △ BCE are equilateral triangles. (1) prove: △ Dah ∽ ECH; (2) if ah: HB = 1:4, find s △ Dah: s △ ECH

(1) In the △ ABC, in △ ABC, in △ ABC, the \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\hb =1：4，∴HB=4AH．∵△ACH∽△CBH，∴CH2=AH•HB=4AH2∵△DAH∽△ECH，∴S△DAH：S△ECH．=AH2：CH2=1：4．

As shown in the figure, ∠ ACB = 90 ° and AC = 2 in △ ABC, the equilateral triangle ACD is made from AC to the right
(1) As shown in Figure 1, rotate the line segment AB around the point a by 60 ° counterclockwise to get the line segment Ab1 and connect with db1, then the line segment with the same length as db1 is___ (write the conclusion directly); (2) as shown in Figure 2, if P is any point on the line BC (not coincident with point C), point P rotates 60 ° counterclockwise around point a to get point Q, and calculate the degree of ∠ ADQ; (3) draw a picture and explore: if P is any point on the line BC (not coincident with point C), point P rotates 60 ° counterclockwise around point a to get point Q, whether there is point P, so that a, C, Q, D are four vertices If it exists, please point out the position of point P and find out the length of PC; if it does not exist, please explain the reason

(1) The line AB is rotated 60 ° anticlockwise around the point a to get the line Ab1, which is connected with db1, then the line with the same length as db1 is BC; so the answer is BC; (2) according to the drawing, AP = AQ, ∠ PAQ = 60 °∫ ACD is an equilateral triangle

As shown in the figure, in △ ABC, ∠ a = 36 °, ABC = 40 °, be bisection ∠ ABC, ∠ e = 18 °, CE bisection ∠ ACD? Why?

The reasons are as follows: ∵ a = 36 °, ABC = 40 °, ∵ BCA = 104 °, ∵ ACD = 76 °. ∵ be = ABC, ∵ CBE = 20 °, ∵ e = 18 °, ∵ BCE = 142 °, ∵ ECA = 38 °, ∵ ECD = 38 °,