I'll try to be more specific

Draw the auxiliary line according to the following steps: (Be sure to look at the picture or you will be confused = =)

First, make the middle point m of BC connect BE, EM and MD.

Do vertical MD from EH to e and H.

Then, in the isosceles triangle EBC, BE=EC, and EM is the midline on the side of BC, so EM is perpendicular to BC.

In an equilateral triangle BCD, DM is the middle line on the side of BC, so DM is perpendicular to BC.

Therefore, BC is perpendicular to EM and DM. So BC is perpendicular to the surface EMD. So BC is perpendicular to any line on this surface. So BC is perpendicular to EH.

Because you are doing EH perpendicular to MD, EH is perpendicular to MD and BC. So EH is perpendicular to the surface BCD.

So the angle between CE and BCD is the ECH angle.

The sine value is EH/EC.

In regular triangle ACD, the center line EC= the root sign of three-halves.

Then go find it

The connection AM has AM=MD

So ME in the isosceles triangle AMD is the midline on the side of AD, so ME is perpendicular to AD.

So there is a right triangle MED, where ME is perpendicular to ED.

So EH is the height next to MD.

It's easy to know that the right triangle MED is similar to the right triangle EHD (I wrote it according to the angle, you have to look at it according to the picture)

So there are: EH/ED=ME/MD

Namely: EH=ME/MD*ED

It is easy to know that in the right triangle MEC

My square = the square of (root number three)-half square = half.

So the root of ME =.

The midline MD in the regular triangle BCD is equal to the root number three; ED= half

Therefore, EH= radical two/radical three * half = radical six.

So sine value =EH/EC= six-sixths root number/three-thirds root number = two-thirds root number.

It's over. I don't know if you understand = =