Answer given is 3 tables which are $R_1(A, B, C, H)$ , $R_2(G, H)$ , $R_3(D, F, A)$
My doubt is if we join $E_1$ and $E_2$ making a new relation $R_{13}(A,G,B,C,H)$ with AG as key
and $R_2(A,D,F)$. So this can be done with only 2 tables as $E_1$ and $E_2$ are totally participating hence no loss of any data and relation $R_2$ with having AD as key can identify $E_2$.