정사각형 내 삼각형의 각도 구하기

문제

한 변의 길이가 6인 정사각형 ABCD가 있습니다. 점 E는 변 CD 위에 있으며 $\overline{CE} = 2\overline{DE}$를 만족합니다. 점 F는 변 BC 위에 있으며 $\overline{BF} = \overline{FC}$를 만족합니다. 이때, $\angle EAF$의 탄젠트 값을 구하시오.

<!-- Vertices of the square -->
<circle cx="50" cy="250" r="3"/>
<text x="40" y="265">A</text>
<circle cx="230" cy="250" r="3"/>
<text x="235" y="265">B</text>
<circle cx="230" cy="70" r="3"/>
<text x="235" y="60">C</text>
<circle cx="50" cy="70" r="3"/>
<text x="40" y="60">D</text>

<!-- Point E -->
<circle cx="110" cy="70" r="3"/>
<text x="100" y="60">E</text>

<!-- Point F -->
<circle cx="230" cy="160" r="3"/>
<text x="235" y="170">F</text>

<!-- Square ABCD -->
<polyline points="50,250 230,250 230,70 50,70 50,250" fill="none"/>

<!-- Lines AE and AF -->
<line x1="50" y1="250" x2="110" y2="70" stroke-dasharray="2 2"/> <!-- AE -->
<line x1="50" y1="250" x2="230" y2="160" stroke-dasharray="2 2"/> <!-- AF -->

<!-- Helper lines for E and F clarity -->
<line x1="50" y1="70" x2="110" y2="70" stroke-width="1" stroke-dasharray="4 2" stroke="#6C63FF"/> <!-- DE segment -->
<line x1="110" y1="70" x2="230" y2="70" stroke-width="1" stroke-dasharray="4 2" stroke="#6C63FF"/> <!-- CE segment -->
<line x1="230" y1="250" x2="230" y2="160" stroke-width="1" stroke-dasharray="4 2" stroke="#6C63FF"/> <!-- BF segment -->
<line x1="230" y1="160" x2="230" y2="70" stroke-width="1" stroke-dasharray="4 2" stroke="#6C63FF"/> <!-- FC segment -->

<!-- Labels for lengths -->
<text x="80" y="88" font-size="12" fill="#6C63FF">2</text> <!-- DE -->
<text x="170" y="88" font-size="12" fill="#6C63FF">4</text> <!-- CE -->
<text x="248" y="205" font-size="12" fill="#6C63FF">3</text> <!-- BF -->
<text x="248" y="115" font-size="12" fill="#6C63FF">3</text> <!-- FC -->

<!-- Angle EAF indication (optional, but helpful) -->
<path d="M 50 250 L 70 242 A 30 30 0 0 1 78 220 L 50 250" fill="none" stroke="#6C63FF" stroke-width="1"/>