정사각형과 직각 관계를 이용한 선분 길이 구하기

문제

그림과 같이 한 변의 길이가 $12$인 정사각형 $\text{ABCD}$가 있습니다. 변 $\text{BC}$ 위에 점 $\text{P}$가 있고, 변 $\text{CD}$ 위에 점 $\text{Q}$가 있습니다. $\text{BP} = 4$ 이고, 선분 $\text{AP}$와 선분 $\text{PQ}$가 서로 수직($\text{AP} \\perp \text{PQ}$)일 때, 선분 $\text{DQ}$의 길이를 구하시오.

graph TD
    A[A] --- B[B]
    B --- C[C]
    C --- D[D]
    D --- A

    B --- P[P] @@ P(200, 400)
    C --- Q[Q] @@ Q(400, 200)

    A --- P
    P --- Q

    style A fill:#fff,stroke:#333,stroke-width:2px,rx:5px,ry:5px;
    style B fill:#fff,stroke:#333,stroke-width:2px,rx:5px,ry:5px;
    style C fill:#fff,stroke:#333,stroke-width:2px,rx:5px,ry:5px;
    style D fill:#fff,stroke:#333,stroke-width:2px,rx:5px,ry:5px;
    style P fill:#fff,stroke:#333,stroke-width:2px,rx:5px,ry:5px;
    style Q fill:#fff,stroke:#333,stroke-width:2px,rx:5px,ry:5px;

    linkStyle 0 stroke:black,stroke-width:1px;
    linkStyle 1 stroke:black,stroke-width:1px;
    linkStyle 2 stroke:black,stroke-width:1px;
    linkStyle 3 stroke:black,stroke-width:1px;
    linkStyle 4 stroke:black,stroke-width:1px;
    linkStyle 5 stroke:black,stroke-width:1px;

    A -- "12" --- B
    B -- "12" --- C
    C -- "12" --- D
    D -- "12" --- A

    B -- "4" --- P
    style P fill:#f9f,stroke:#333,stroke-width:2px;

    link A -- P
    link P -- Q

    subgraph Hidden Right Angle
        direction LR
        Z(( )) -- Z1(( ))
        Z1 -- Z2(( ))
    end
    style Z visibility:hidden;
    style Z1 visibility:hidden;
    style Z2 visibility:hidden;

    Z --- A
    Z1 --- B
    Z2 --- C
    Z --- D

    
    classDef rightangle fill:#fff,stroke:#000,stroke-width:0px;
    classDef rightangle_marker fill:#000;

    class B rightangle_marker;
    class C rightangle_marker;
    class angleAPQ rightangle_marker;