삼각형의 각 이등분선과 평행선을 이용한 길이 추론

문제

다음 그림과 같이 $\triangle ABC$에서 $\angle BAC$의 이등분선이 변 $BC$와 만나는 점을 $D$라 하자. 점 $C$를 지나고 선분 $AD$에 평행한 직선이 변 $BA$의 연장선과 만나는 점을 $E$라 할 때, $AC = 8$ cm 이다. 다음 중 항상 옳은 것은?

\begin{center}
\begin{tikzpicture}[scale=1]
    \coordinate (A) at (0,3);
    \coordinate (B) at (-2,-1);
    \coordinate (C) at (3,-1);

    \draw (A) node[above] {$A$} -- (B) node[left] {$B$} -- (C) node[right] {$C$} -- cycle;

    % Angle bisector AD
    \coordinate (AD_interm) at ($(B)!0.5!(C)$); % D could be anywhere on BC
    % To ensure D is on BC and AD is a bisector visually accurate, we need angle. 
    % Let's calculate D more accurately by ratios AB:AC = BD:DC 
    % But this is not needed for the diagram's purpose for the problem. 
    % A simplified D on BC is sufficient.
    \coordinate (D) at ($ (B)!0.6!(C) $); % Just a point on BC
    \draw[thick] (A) -- (D) node[below] {$D$};
    \pic [draw, angle radius=0.7cm, "$\\alpha$"] {angle = B--A--D};
    \pic [draw, angle radius=0.7cm, "$\\alpha$"] {angle = D--A--C};

    % Extend BA to E
    \coordinate (E) at ($ (A) + 1.2*(A) - 1.2*(B) $); % E on BA extended
    \draw[dashed] (B) -- (E) node[left] {$E$};

    % Line CE parallel to AD
    \path (A) -- (D);
    \coordinate (AD_vec) at ($(D)-(A)$);
    \coordinate (E_new) at ($ (C) + (AD_vec)*1.5 $); % A point in the direction of AD from C
    \draw[thick] (C) -- (E);
    
    % Parallel line marks
    \draw[->, gray] ($(A)!0.5!(D)$) + (0.1,0.1) -- + (0.3,0.3);
    \draw[->, gray] ($(C)!0.5!(E)$) + (0.1,0.1) -- + (0.3,0.3);

\end{tikzpicture}
\end{center}