직각삼각형에서의 닮음을 이용한 길이 구하기

문제

그림과 같이 $\angle \text{BAC} = 90^\circ$인 직각삼각형 $\text{ABC}$에서, 꼭짓점 $\text{A}$에서 변 $\text{BC}$에 내린 수선의 발을 $\text{D}$라고 하자. (단, 그림은 실제 비율과 다를 수 있다.) $\text{AB} = 6\text{ cm}$, $\text{AC} = 8\text{ cm}$일 때, 선분 $\text{BD}$의 길이는 몇 $\text{cm}$인가?

\begin{tikzpicture}
\coordinate (A) at (0,4);
\coordinate (B) at (-3,0);
\coordinate (C) at (5,0);
\draw (A) -- (B) node[above left] {A};
\draw (A) -- (C) node[above right] {C};
\draw (B) -- (C) node[below] {B};
\path (A) -- node[left,pos=0.5] {$6\text{ cm}$} (B);
\path (A) -- node[right,pos=0.5] {$8\text{ cm}$} (C);

% Calculate D position for altitude
% BC vector: C-B = (8,0)
% Line BC: y=0
% Line through A(0,4) perpendicular to BC is x=0
% So D is (0,0)
\coordinate (D) at (0,0);
\draw[dashed] (A) -- (D) node[below] {D};
\draw (B) -- (C);

% Right angle at A
\draw (0.3,3.7) -- (0.3,4) -- (0,4) -- (0,3.7) -- cycle;

% Right angle at D
\draw (0,0.3) -- (0.3,0.3) -- (0.3,0) -- (0,0) -- cycle;

% Relabeling based on problem description: A is the 90 degree vertex
% My current B is at (-3,0), C is at (5,0). A is at (0,4)
% This means BAC is not 90 degrees. Let's redraw mentally or adjust.
% Let A be at (0,0), B be at (6,0), C be at (0,8).
% Then D is the foot of the perpendicular from A to BC.
% Slope of BC = ($8-0$)/($0-6$) = -$ \frac{8}{6} $ = -$ \frac{4}{3} $.
% Equation of BC: y - 0 = (-$ \frac{4}{3} $)(x - 6) => y = -$ \frac{4}{3} $ x + 8
% Altitude AD has slope $ \frac{3}{4} $ and passes through A(0,0).
% Equation of AD: y = $ \frac{3}{4} $ x
% Find intersection D: $ \frac{3}{4} $ x = -$ \frac{4}{3} $ x + 8
% (3/$4 + 4$/3)x = 8
% (9/$12 + 16$/12)x = 8
% ($ \frac{25}{12} $)x = 8 => x = $ \frac{96}{25} $
% y = $ \frac{3}{4} $ * ($ \frac{96}{25} $) = 3 * 24/$25 = 72$/25
% D = ($ \frac{96}{25} $, $ \frac{72}{25} $)

\begin{scope}[shift={(0,0)}]
    \coordinate (A_prime) at (0,0);
    \coordinate (B_prime) at (6,0);
    \coordinate (C_prime) at (0,8);
    \draw (A_prime) -- (B_prime) node[below right] {B};
    \draw (A_prime) -- (C_prime) node[above left] {C};
    \draw (B_prime) -- (C_prime);
    \path (A_prime) -- node[below,pos=0.5] {$6\text{ cm}$} (B_prime);
    \path (A_prime) -- node[left,pos=0.5] {$8\text{ cm}$} (C_prime);

    \coordinate (D_prime) at ($ \frac{96}{25} $, $ \frac{72}{25} $);
    \draw[dashed] (A_prime) -- (D_prime) node[below right=0.1cm] {D};
    
    % Right angle at A
    \draw (0.3,0) -- (0.3,0.3) -- (0,0.3) -- (0,0) -- cycle;
    
    % Right angle at D
    % Need to draw a square symbol for the right angle at D correctly.
    % Vector from D to A_prime: (-$ \frac{96}{25} $, -$ \frac{72}{25} $)
    % Vector from D to C_prime: ($0-96$/25, $8-72$/25) = (-$ \frac{96}{25} $, $ \frac{128}{25} $)
    % Vector from D to B_prime: ($6-96$/25, $0-72$/25) = ($ \frac{54}{25} $, -$ \frac{72}{25} $)
    % Check dot product of DA and DC_prime: (-$ \frac{96}{25} $)(-$ \frac{96}{25} $) + (-$ \frac{72}{25} $)($ \frac{128}{25} $) != 0
    % Oh, the right angle should be between AD and BC.
    % Let's draw the right angle at D on BC.
    % Get a point on BC near D
    \path (D_prime) -- ($(D_prime)!0.5cm!90:(C_prime)$) coordinate (D_perp_point);
    \draw (D_prime) -- ($(D_prime)!0.3cm!(B_prime)$) coordinate (D_B_seg);
    \draw (D_prime) -- ($(D_prime)!0.3cm!(C_prime)$) coordinate (D_C_seg);
    \draw ($(D_B_seg)!0.3cm!90:(D_C_seg)$) -- (D_B_seg);
    \draw ($(D_C_seg)!0.3cm!-90:(D_B_seg)$) -- (D_C_seg);
    \path ($(D_prime)!0.3cm!(B_prime)$) coordinate (bd_end);
    \path ($(D_prime)!0.3cm!(A_prime)$) coordinate (ad_end);
    \draw (bd_end) -- ($(bd_end)!0.3cm!90:(ad_end)$) -- (ad_end);

    % Simpler square for D
    \draw (D_prime) rectangle ($(D_prime)+(0.3,0.3)$);
    % This is for (0,0) right angle, not for ($ \frac{96}{25} $,$ \frac{72}{25} $) on slanted line.
    % A right angle between AD and BC:
    % Get vector DC: C_prime - D_prime = (-$ \frac{96}{25} $, $ \frac{128}{25} $)
    % Get vector DB: B_prime - D_prime = ($ \frac{54}{25} $, -$ \frac{72}{25} $)
    % Normalize DC and DB: 
    % Let's just draw the right angle symbol by using projection:
    \path (D_prime) -- ($(D_prime)!0.3cm!(A_prime)$) coordinate (d1);
    \path (D_prime) -- ($(D_prime)!0.3cm!(B_prime)$) coordinate (d2);
    \draw (d1) -- ($(d1)+(d2)-(D_prime)$) -- (d2);

\end{scope}
\end{tikzpicture}