SAMC

Let $AB=c$, $BC=a$, $CA=b$, $B{A}^{\prime }=x$, $C{B}^{\prime }=y$, $A{C}^{\prime }=z$. The condition in the problem $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=\frac{3}{2}$ is equivalent to $$\begin{array}{c}bcx+cay+abz=\frac{3}{2}abc\end{array}\text{\hspace{1em}(1)}$$ From Ceva's Theorem, $$\frac{x}{a-x}\cdot \frac{y}{b-y}\cdot \frac{z}{c-z}=1,$$ so $$\begin{array}{c}xyz=(a-x)(b-y)(c-z)=abc-(bcx+cay+abz)\\ +(axy+byz+czx)-xyz\end{array}$$ or $$2xyz=\frac{1}{4}abc+\frac{3}{4}abc-(bcx+cay+abz)+(axy+byz+czx).$$ The (1) is equivalent to $$2xyz=\frac{1}{4}abc-\frac{1}{2}(bcx+cay+abz)+(axy+byz+czx),$$ which can be written as $$8xyz-4(axy+byz+czx)+2(bcx+cay+abz)-abc=0$$ so $(2x-a)(2y-b)(2z-c)=0$.

This means that at least one of the segments $A{A}^{\prime }$, $B{B}^{\prime }$, $C{C}^{\prime }$ is a median in triangle $ABC$ and the conclusion follows.

Solution 2:

Denote $$x=\frac{A^{\prime }C}{A^{\prime }B},y=\frac{B^{\prime }A}{B^{\prime }C},z=\frac{C^{\prime }B}{C^{\prime }A}.$$ From Ceva's Theorem, we have $xyz=1$. The condition in the problem is equivalent to $$\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=\frac{3}{2}$$ That is $$\begin{array}{c}2[(x+1)(y+1)+(y+1)(z+1)+(z+1)(x+1)]\\ =3(x+1)(y+1)(z+1)\end{array}$$ We obtain $$\begin{array}{c}2(xy+yz+zx)+4(x+y+z)+6=3xyz+3(xy+yz+zx)\\ +3(x+y+z)+3\end{array}$$ hence $$3xyz-(x+y+z)+(xy+yz+zx)-3=0$$ that is $$1-(x+y+z)+(xy+yz+zx)-1=0.$$ It follows $$xyz-(x+y+z)+(xy+yz+zx)-1=0$$ hence $$(x-1)(y-1)(z-1)=0$$ and the conclusion follows.