11. 如图18-37,四边形ABCD是菱形,$AE⊥BC$于点E,$AF⊥CD$于点F.
(1)求证:$△ABE\cong △ADF$;
(2)若$AE=4,CF=2$,求菱形的边长.

(1) ∵ 四边形 $ABCD$ 是菱形，∴ $AB = BC = CD = AD$，$∠B = ∠D$。∵ $AE ⊥ BC$，$AF ⊥ CD$，∴ $∠AEB = ∠AFD = 90°$。在 $△ABE$ 和 $△ADF$ 中，$\begin{cases}∠AEB = ∠AFD\\∠B = ∠D\\AB = AD\end{cases}$，∴ $△ABE ≌ △ADF (AAS)$。
(2) 设菱形的边长为 $x$。∵ $AB = CD = x$，$CF = 2$，∴ $DF = x - 2$。∵ $△ABE ≌ △ADF$，∴ $BE = DF = x - 2$。在 $Rt△ABE$ 中，根据勾股定理，得 $AE² + BE² = AB²$，即 $4² + (x - 2)² = x²$，解得 $x = 5$，∴ 菱形的边长是 5。

12. 在矩形ABCD中,BD是对角线,$AE⊥BD$于点E,$CF⊥BD$于点F.

(1)如图18-38①,求证:$AE=CF$;
(2)如图18-38②,当$∠ADB=30^{\circ }$时,连接AF,CE,在不添加任何辅助线的情况下,请直接写出图18-38②中四个三角形,使写出的每个三角形的面积都等于矩形ABCD面积的$\frac {1}{8}$.

(1) ∵ 四边形 $ABCD$ 是矩形，∴ $AB = CD$，$AB // CD$，$AD // BC$，∴ $∠ABE = ∠CDF$。∵ $AE ⊥ BD$ 于点 $E$，$CF ⊥ BD$ 于点 $F$，∴ $∠AEB = ∠CFD = 90°$。在 $△ABE$ 和 $△CDF$ 中，$\begin{cases}∠ABE = ∠CDF\\∠AEB = ∠CFD\\AB = CD\end{cases}$，∴ $△ABE ≌ △CDF (AAS)$，∴ $AE = CF$。
(2) $△ABE$，$△CDF$，$△BCE$，$△ADF$ 都等于矩形 $ABCD$ 面积的 $\frac{1}{8}$。提示：∵ $AD // BC$，∴ $∠CBD = ∠ADB = 30°$。∵ $∠ABC = 90°$，∴ $∠ABE = 60°$。∵ $AE ⊥ BD$，∴ $∠BAE = 30°$，∴ $BE = \frac{1}{2}AB$，$AE = \frac{1}{2}AD$，∴ $S_{△ABE} = \frac{1}{2}BE × AE = \frac{1}{2} × \frac{1}{2}AB × \frac{1}{2}AD = \frac{1}{8}AB × AD = \frac{1}{8}S_{矩形ABCD}$。∵ $△ABE ≌ △CDF$，∴ $S_{△CDF} = \frac{1}{8}S_{矩形ABCD}$。作 $EG ⊥ BC$ 于点 $G$，∵ $∠CBD = 30°$，∴ $EG = \frac{1}{2}BE = \frac{1}{2} × \frac{1}{2}AB = \frac{1}{4}AB$，∴ $S_{△BCE} = \frac{1}{2}BC × EG = \frac{1}{2}BC × \frac{1}{4}AB = \frac{1}{8}BC × AB = \frac{1}{8}S_{矩形ABCD}$。同理，$S_{△ADF} = \frac{1}{8}S_{矩形ABCD}$。