작도와 합동 - 숨겨진 진실 찾기

문제

다음 그림과 같이 $\triangle \text{ABC}$에서 변 $\text{AB}$ 위에 점 $\text{D}$를 $\overline{\text{AD}} = \overline{\text{AC}}$가 되도록 잡는다. $\angle \text{BAC}$의 이등분선이 변 $\text{BC}$와 만나는 점을 $\text{E}$라 하고, 선분 $\overline{\text{AE}}$와 선분 $\overline{\text{DC}}$의 교점을 $\text{P}$라고 할 때, 다음 중 항상 참이라고 할 수 없는 것은?

\begin{tikzpicture}[scale=1] % Define coordinates for a general triangle ABC \coordinate (A) at (0,4); \coordinate (B) at (-4,0); \coordinate (C) at (5,0);

% Draw triangle ABC \draw (A) node[above] {A} -- (B) node[left] {B} -- (C) node[right] {C} -- cycle;

% Point D on AB such that AD = AC % Calculate AC length \pgfmathparse{sqrt(($5-0$)^2 + ($0-4$)^2)} \let\ACLength\pgfmathresult % Calculate vector AB \pgfmathparse{(-$4-0$)} \let\ABx\pgfmathresult \pgfmathparse{($0-4$)} \let\ABy\pgfmathresult % Calculate length of AB \pgfmathparse{sqrt(\ABx^2 + \ABy^2)} \let\ABLength\pgfmathresult % Calculate D point using proportional vector from A to B \pgfmathparse{0 + \ACLength * \ABx / \ABLength} \let\Dx\pgfmathresult \pgfmathparse{4 + \ACLength * \ABy / \ABLength} \let\Dy\pgfmathresult \coordinate (D) at (\Dx,\Dy); \fill (D) circle (1.5pt) node[left] {D};

% Draw line AD and AC with tick marks to show AD=AC \draw[thick] (A) -- (D); \draw[thick] (A) -- (C); \draw[thick] (D) -- (C); % Line segment DC

% Angle bisector AE of BAC % Calculate angle bisector direction \pgfmathparse{atan2(($0-4$),(-$4-0$))} \let\angleAB\pgfmathresult % Angle of vector AB relative to A \pgfmathparse{atan2(($0-4$),($5-0$))} \let\angleAC\pgfmathresult % Angle of vector AC relative to A \pgfmathparse{(\angleAB + \angleAC)/2} \let\anglebisect\pgfmathresult \pgfmathparse{$0 + 5$*cos(\anglebisect)} \let\Ex\pgfmathresult % Extend for intersection \pgfmathparse{$4 + 5$*sin(\anglebisect)} \let\Ey\pgfmathresult \coordinate (AE_dir) at (\Ex,\Ey);

% Find intersection of AE_dir and BC to get E \path[name path=AE_path] (A) -- ($(A)!2.5!(AE_dir)$); \path[name path=BC_path] (B) -- (C); \path [name intersections={of=AE_path and BC_path, by=E}]; \fill (E) circle (1.5pt) node[below right] {E};

% Draw angle bisector AE \draw (A) -- (E);

% Intersection P of AE and DC \path[name path=AE_seg] (A) -- (E); \path[name path=DC_seg] (D) -- (C); \path [name intersections={of=AE_seg and DC_seg, by=P}]; \fill (P) circle (1.5pt) node[above right] {P};

% Mark AD=AC with simple ticks \draw[line width=0.5mm] ($(A)!0.15!(D)$) -- ($(A)!0.15!(D)+(-0.05,0.05)$); \draw[line width=0.5mm] ($(A)!0.15!(D)$) -- ($(A)!0.15!(D)+(0.05,-0.05)$); \draw[line width=0.5mm] ($(A)!0.15!(C)$) -- ($(A)!0.15!(C)+(-0.05,0.05)$); \draw[line width=0.5mm] ($(A)!0.15!(C)$) -- ($(A)!0.15!(C)+(0.05,-0.05)$);

% Mark angles for AE bisects BAC (single arc for each) - difficult for general, so omit visual if needed. % Let's use two arcs for DAC and CAE to indicate bisection % Approximate angles visually \draw (A) + (240:0.4cm) arc (240:260:0.4cm); \draw (A) + (270:0.4cm) arc (270:290:0.4cm);

\end{tikzpicture}