Extend the intersection of GP and DC at h point

It is easy to prove DC//AB//GF, GF=BG and DC=BC according to the nature of diamonds.

Therefore, it is easy to get that all triangles DPH are equal to triangles FPG, and pH = PG, DH = GF = BG.

So DC-DH=BC-BG, that is, HC=GC, and because HP=GP and PC=PC.

Through the superposition of three isosceles triangles, it is easy to get that PC is perpendicular to PG, angle PCG= 1/2 angle DCB=60 degrees.

So PC/PG= 1/ root number 3.

The answer to question 2 is the same as question 1, but the proof method is slightly different.

Prolonged GP and AD intersect at h point, continuous CH, CG

According to the nature of the diamond, BG=FG, angle GBF= 1/2 angle GBE = 60 degrees, so angle CBG= 180 degrees-angle ABC- angle GBF=60 degrees.

It is easy to prove that triangle DHP is all equal to triangle FGP, and DH=FG=BG is obtained, and because DC=BC, angle HDC= angle GBC=60 degrees.

As can be seen from the angle, the triangle HDC is equal to the triangle GBC, so CH=CG,

According to the coincidence theorem of three lines of isosceles triangle, CP is perpendicular to PG, from which the angle PCG=60 degrees can be obtained.

So in the right triangle PCG, PC/PG= 1/ root number 3.

(Because it is not a word editing software, some symbols cannot be expressed mathematically. )