5. 如图18-22,在菱形$ABCD$中,$E$,$F是对角线AC$上两点,连接$DE$,$DF$,$∠ADF = ∠CDE$. 求证：$AE = CF$.
证明：∵ 四边形ABCD是菱形，∴ ，∴ 。∵ ∠ADF = ∠CDE，∴ ，∴ 。在△DAE和△DCF中，$\left\{\begin{array}{l}\angle DAC = \angle DCA,\\DA = DC,\\\angle ADE = \angle CDF,\end{array}\right.$ ∴ ，∴ AE = CF。

∵ 四边形ABCD是菱形，∴ DA = DC，∴ ∠DAC = ∠DCA。∵ ∠ADF = ∠CDE，∴ ∠ADF - ∠EDF = ∠CDE - ∠EDF，∴ ∠ADE = ∠CDF。在△DAE和△DCF中，$\left\{\begin{array}{l}\angle DAC = \angle DCA,\\DA = DC,\\\angle ADE = \angle CDF,\end{array}\right.$ ∴ △DAE ≌ △DCF(ASA)，∴ AE = CF。

6. 如图18-23,在$□ ABCD$中,$AC$,$BD交于点O$,点$E$,$F在AC$上,$AE = CF$.
(1)求证：四边形$EBFD$是平行四边形；

(2)若$∠BAC = ∠DAC$,求证：四边形$EBFD$是菱形.

(1)∵ 在▱ABCD中，AC，BD交于点O，∴ OA = OC，OB = OD。∵ AE = CF，∴ OE = OF，∴ 四边形EBFD是平行四边形。 (2)∵ 四边形ABCD是平行四边形，∴ AB//DC，∴ ∠BAC = ∠DCA。∵ ∠BAC = ∠DAC，∴ ∠DCA = ∠DAC，∴ DA = DC，∴ 平行四边形ABCD为菱形，∴ DB⊥EF，∴ 平行四边形EBFD是菱形，∴ 四边形EBFD是菱形。