1/179/232/QUIZ1.376001090474666.xml:]]></question><correct><![CDATA[$\rm \angle BAC= \angle BDC$ , $\rm \overline {BC}$ , SSS]]></correct><clue item="0"><![CDATA[대응하는 세 변의 길이가 각각 같을 때 ( SSS 합동)]]></clue><solution item="0"><![CDATA[\titledbox{가정 : $\rm \overline {AB} = \overline {DB} ,~ \overline {AC} = \overline {DC}$par결론 : $\rm \angle BAC= \angle BDC$par증명 : $\rm \triangle ABC$와 $\rm\triangle BAD$에서par$\rm \overline {AB} = \overline {DB}$  (가정)par$\rm \overline {AC} = \overline {DC}$  (가정)par$\rm \overline {BC}$ 는 공통이므로par$\rm \triangle ABC \equiv \triangle DBC$  ( SSS 합동)\string\$\rm \therefore ~ \angle BAC= \angle BDC$}
1/179/237/QUIZ1.377531740774731.xml:\end{array}$ par  따라서 $\rm \overline {AC} = \dfrac {\rm \overline {CH}} {cos 30 \text{\textdegree}}$ $= \dfrac {10} {\sqrt {\dfrac {3} {2}}} = \dfrac {20} {\sqrt {3}}$ $= \dfrac {20 \sqrt {3}} {3} ( \rm cm)$ 이다.
1/179/237/QUIZ1.377599815662372.xml:\end{array}$ par  따라서 $\rm \overline {AC} = \dfrac {\rm \overline {CH}} {cos 30 \text{\textdegree}}$ $= \dfrac {10} {\sqrt {\dfrac {3} {2}}} = \dfrac {20} {\sqrt {3}}$ $= \dfrac {20 \sqrt {3}} {3} ( \rm cm)$ 이다.
1/179/237/QUIZ1.377705823983315.xml:\end{array}$ par  따라서 $\rm \overline {AC} = \dfrac {\rm \overline {CH}} {cos 30 \text{\textdegree}}$ $= \dfrac {10} {\sqrt {\dfrac {3} {2}}} = \dfrac {20} {\sqrt {3}}$ $= \dfrac {20 \sqrt {3}} {3} ( \rm cm)$ 이다.
1/179/238/QUIZ1.377983313801739.xml:]]></question><correct><![CDATA[$\rm \angle BAC= \angle BDC$ , $\rm \overline {BC}$ , SSS]]></correct><clue item="0"><![CDATA[대응하는 세 변의 길이가 각각 같을 때 ( SSS 합동)]]></clue><solution item="0"><![CDATA[\titledbox{가정 : $\rm \overline {AB} = \overline {DB} ,~ \overline {AC} = \overline {DC}$par결론 : $\rm \angle BAC= \angle BDC$par증명 : $\rm \triangle ABC$와 $\rm\triangle BAD$에서par$\rm \overline {AB} = \overline {DB}$  (가정)par$\rm \overline {AC} = \overline {DC}$  (가정)par$\rm \overline {BC}$ 는 공통이므로par$\rm \triangle ABC \equiv \triangle DBC$  ( SSS 합동)\string\$\rm \therefore ~ \angle BAC= \angle BDC$}
1/179/240/QUIZ1.378466449697862.xml:]]></question><correct><![CDATA[$\rm \angle BAC= \angle BDC$ , $\rm \overline {BC}$ , SSS]]></correct><clue item="0"><![CDATA[대응하는 세 변의 길이가 각각 같을 때 ( SSS 합동)]]></clue><solution item="0"><![CDATA[\titledbox{가정 : $\rm \overline {AB} = \overline {DB} ,~ \overline {AC} = \overline {DC}$par결론 : $\rm \angle BAC= \angle BDC$par증명 : $\rm \triangle ABC$와 $\rm\triangle BAD$에서par$\rm \overline {AB} = \overline {DB}$  (가정)par$\rm \overline {AC} = \overline {DC}$  (가정)par$\rm \overline {BC}$ 는 공통이므로par$\rm \triangle ABC \equiv \triangle DBC$  ( SSS 합동)\string\$\rm \therefore ~ \angle BAC= \angle BDC$}
1/179/240/QUIZ1.378588126059245.xml:]]></question><correct><![CDATA[$\rm \angle BAC= \angle BDC$ , $\rm \overline {BC}$ , SSS]]></correct><clue item="0"><![CDATA[대응하는 세 변의 길이가 각각 같을 때 ( SSS 합동)]]></clue><solution item="0"><![CDATA[\titledbox{가정 : $\rm \overline {AB} = \overline {DB} ,~ \overline {AC} = \overline {DC}$par결론 : $\rm \angle BAC= \angle BDC$par증명 : $\rm \triangle ABC$와 $\rm\triangle BAD$에서par$\rm \overline {AB} = \overline {DB}$  (가정)par$\rm \overline {AC} = \overline {DC}$  (가정)par$\rm \overline {BC}$ 는 공통이므로par$\rm \triangle ABC \equiv \triangle DBC$  ( SSS 합동)\string\$\rm \therefore ~ \angle BAC= \angle BDC$}
1/179/240/QUIZ1.378610630160238.xml:]]></question><correct><![CDATA[$\rm \angle BAC= \angle BDC$ , $\rm \overline {BC}$ , SSS]]></correct><clue item="0"><![CDATA[대응하는 세 변의 길이가 각각 같을 때 ( SSS 합동)]]></clue><solution item="0"><![CDATA[\titledbox{가정 : $\rm \overline {AB} = \overline {DB} ,~ \overline {AC} = \overline {DC}$par결론 : $\rm \angle BAC= \angle BDC$par증명 : $\rm \triangle ABC$와 $\rm\triangle BAD$에서par$\rm \overline {AB} = \overline {DB}$  (가정)par$\rm \overline {AC} = \overline {DC}$  (가정)par$\rm \overline {BC}$ 는 공통이므로par$\rm \triangle ABC \equiv \triangle DBC$  ( SSS 합동)\string\$\rm \therefore ~ \angle BAC= \angle BDC$}
1/179/241/QUIZ1.378724614352345.xml:]]></question><correct><![CDATA[$\rm \angle BAC= \angle BDC$ , $\rm \overline {BC}$ , SSS]]></correct><clue item="0"><![CDATA[대응하는 세 변의 길이가 각각 같을 때 ( SSS 합동)]]></clue><solution item="0"><![CDATA[\titledbox{가정 : $\rm \overline {AB} = \overline {DB} ,~ \overline {AC} = \overline {DC}$par결론 : $\rm \angle BAC= \angle BDC$par증명 : $\rm \triangle ABC$와 $\rm\triangle BAD$에서par$\rm \overline {AB} = \overline {DB}$  (가정)par$\rm \overline {AC} = \overline {DC}$  (가정)par$\rm \overline {BC}$ 는 공통이므로par$\rm \triangle ABC \equiv \triangle DBC$  ( SSS 합동)\string\$\rm \therefore ~ \angle BAC= \angle BDC$}
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