这道题怎么做？---**1. Textual Information:** * **Question Stem Title:** 题干 (Question Stem) * **Question Stem Content:** 已知 $\Box ABCD$, $O$ 是对角线 $AC$ 与 $BD$ 的交点, $OE$ 是 $\triangle ABC$ 的中位线, 联结 $AE$ 并延长与 $DC$ 的延长线交于点 $F$, 联结 $BF$. 求证: 四边形 $ABFC$ 是平行四边形. (Translation: Given parallelogram $ABCD$, $O$ is the intersection of diagonals $AC$ and $BD$. $OE$ is the midsegment of $\triangle ABC$. Connect $AE$ and extend it to intersect the extension of $DC$ at point $F$. Connect $BF$. Prove that quadrilateral $ABFC$ is a parallelogram.) **2. Chart/Diagram Description:** * **Type:** Geometric figure. * **Main Elements:** * **Shapes:** * A parallelogram is depicted, labeled $ABCD$, with vertices $A$ at the top-left, $B$ at the bottom-left, $C$ at the bottom-right, and $D$ at the top-right. * Several triangles are formed within and around the parallelogram, including $\triangle ABC$, $\triangle ADC$, $\triangle ABD$, $\triangle BCD$, $\triangle AOE$, $\triangle COE$, $\triangle ABE$, $\triangle AEC$, $\triangle EBC$, $\triangle ABF$, $\triangle ACF$, $\triangle BCF$. * A quadrilateral $ABFC$ is formed, which is the subject of the proof. * **Points:** * Vertices of the parallelogram are $A$, $B$, $C$, $D$. * Point $O$ is the intersection of the diagonals $AC$ and $BD$. * Point $E$ is located on the side $BC$ of the parallelogram. * Point $F$ is an external point, formed by the intersection of the extended line segment $AE$ and the extended line segment $DC$. * **Lines/Segments:** * The sides of the parallelogram are $AB$, $BC$, $CD$, $DA$. * The diagonals of the parallelogram are $AC$ and $BD$. * Line segment $OE$ connects point $O$ (intersection of diagonals) to point $E$ (on $BC$). * Line segment $AE$ is drawn, and then extended past $E$ to point $F$. * Line segment $DC$ is extended past $C$ to point $F$. * Line segment $BF$ connects point $B$ to point $F$. * Other visible segments include $AO$, $OC$, $BO$, $OD$, $BE$, $EC$. * **Relative Position and Direction:** * $ABCD$ is a parallelogram, implying $AB \parallel DC$, $AD \parallel BC$, $AB=DC$, $AD=BC$. * $O$ is the midpoint of $AC$ and $BD$ because it's the intersection of the diagonals of a parallelogram. * $E$ is the midpoint of $BC$ because $OE$ is described as the midsegment of $\triangle ABC$, and $O$ is the midpoint of $AC$ (from the parallelogram property), and $E$ is shown on $BC$. * Points $A$, $E$, $F$ are collinear (lie on the same straight line). * Points $D$, $C$, $F$ are collinear (lie on the same straight line), with $C$ between $D$ and $F$. * The figure shows the extension of $AE$ and $DC$ meeting at $F$, indicating $F$ is outside the original parallelogram $ABCD$.