The first topic is A. We can know that the triangle area changes regularly when the moving point is moving. When moving on BC, the area increases continuously, when moving on CD, the area remains unchanged, and when moving on DA, the area decreases. As can be seen from the figure on the right, when X=4, the area does not increase, that is to say, BC=4. During the process from 4 to 9, the area of the triangle remains unchanged, so the length of CD is 5. Therefore, the area of △ABC is 0.5×4×5= 10.

The second question 1235 is correct. ① It can be proved that △ADC is all equal to △BEC by "edges and corners". Yes, the corresponding edges are equal, and AD=BE.

② You need to use the congruence condition of ③, and then we will talk about it.

③ From the congruence condition of ①, there can be ∠DAE=∠CBE, ∠BCA=∠BCD and AD=BC in the regular triangle. So △BQC is equal to △APC. The corresponding edges are equal, AP=BQ.

Now say ②, from the congruence of these two groups of triangles, we can have, AD=BE, AP=BQ. So AD-AP=BE-BQ, which means PD=QE. From the congruence condition in ①, there can be ∠ADC=∠BEC and ∠CDE=∠DCB, and it can be concluded that △ECQ is equal to △DCP, so there is CP=CQ. And ∠ pcq = 60, so △PCQ is that all three internal angles are regular triangles.

4 is wrong. There is DE=DC in the regular triangle. If DE=DP, and DC=DP happens, then △DPC is a regular triangle, which is obviously impossible because ∠ ADC < 60.

⑤ From the congruence condition in ①, we know that ∠DAE=∠EBC, so ∠ AOB = ∠ OAE+∠ BEC = ∠ EBC+∠ BEC = ∠ ACB = 60.

I'm afraid you're tired of watching too much. Look at these two first, and I'll drink some water and continue counting.

The third question, I wrote it down directly, deliberately in big letters. If you can't see clearly, feel free to ask me.