smc

Cross-multiplying gives: $(a+b)(a+d)=(b+c)(c+d)$ $a^2+ad+ab+bd=bc+bd+c^2+cd$ $a^2+ad+ab-bc-c^2-cd=0$ $a(a+b+d)-c(b+c+d)=0$ This looks close to turning into option C, but we don't have a $c$ term in the first parentheses, and we don't have an $a$ term in the second parentheses to allow us to complete the factorization. However, if we both add $ac$ and subtract $ac$ on the LHS, we get: $a(a+b+d)+ac-c(b+c+d)-ca=0$ $a(a+b+d+c)-c(b+c+d+a)=0$ $(a-c)(a+b+c+d)=0$ This is equivalent to $\boxed{\text{either }a=c\text{ or }a+b+c+d=0\text{, or both}}$