이등변삼각형 속 숨겨진 합동 찾기

문제

그림과 같이 $AB=AC$인 이등변삼각형 $ABC$에서, 변 $AB$ 위의 점 $D$와 변 $AC$ 위의 점 $E$를 각각 잡는다. 선분 $CD$와 선분 $BE$는 점 $F$에서 만난다. $BD = CE$일 때, 다음 중 옳지 않은 것은?

\begin{center}
\begin{tikzpicture}
\coordinate (A) at (0,3);
\coordinate (B) at (-2,0);
\coordinate (C) at (2,0);
\draw (A) node[above] {$A$} -- (B) node[left] {$B$} -- (C) node[right] {$C$} -- cycle;

% D on AB, E on AC such that BD=CE
% Let D be 40% along AB from B towards A
\coordinate (D) at ($0.6*(B) + 0.4*(A)$);
\node at (D) [left] {$D$};

% Let E be 40% along AC from C towards A
\coordinate (E) at ($0.6*(C) + 0.4*(A)$);
\node at (E) [right] {$E$};

% Draw CD and BE
\draw (C) -- (D);
\draw (B) -- (E);

% F is intersection of CD and BE
\coordinate (F) at (intersection of C--D and B--E);
\node at (F) [above right] {$F$};

% Add BD and CE length marks (one short line for each to show equality)
\draw [line width=0.5pt, black] ($(B)!0.1!(D)$) -- ($(B)!0.2!(D)$);
\draw [line width=0.5pt, black] ($(C)!0.1!(E)$) -- ($(C)!0.2!(E)$);

% Add AB and AC length marks (two short lines for each to show equality)
\draw [line width=0.5pt, black] ($(A)!0.1!(B)$) -- ($(A)!0.15!(B)$);
\draw [line width=0.5pt, black] ($(A)!0.2!(B)$) -- ($(A)!0.25!(B)$);
\draw [line width=0.5pt, black] ($(A)!0.1!(C)$) -- ($(A)!0.15!(C)$);
\draw [line width=0.5pt, black] ($(A)!0.2!(C)$) -- ($(A)!0.25!(C)$);
\end{tikzpicture}
\end{center}