14. 在$△ ABC$中，$∠ ABC$的平分线$BD$与$AC$相交于点$D$，$DE⊥ AB$，垂足为$E$.
(1)如图$1$，若$∠ ABC=90^{\circ}$，则$∠ EDB$的度数为.
(2)如图$2$，若$△ ABC$是锐角三角形，过点$E$作$EF// BC$，交$AC$于点$F$. 依题意补全图形，用等式表示$∠ FED$、$∠ EDB$与$∠ ABC$之间的数量关系并证明.
(3)若$△ ABC$是钝角三角形，其中$90^{\circ}＜∠ BAC＜180^{\circ}$. 过点$E$作$EF// BC$，交直线$AC$于点$F$，依题意画出图形，并直接写出$∠ FED$、$∠ EDB$与$∠ ABC$之间的数量关系.

14. (1)45° 解析:
∵∠AED = ∠ABC = 90°,
∴ED//BC,
∴∠EDB = ∠DBC = $\frac{1}{2}$∠ABC = $\frac{1}{2}$×90° = 45°.
(2)补全图形如图1所示.∠FED、∠EDB与∠ABC之间的数量关系为2(∠EDB - ∠FED)=∠ABC.证明如下:
∵EF//BC,
∴∠AEF = ∠ABC.又
∵∠EBD + ∠EDB = ∠AEF + ∠FED,
∴$\frac{1}{2}$∠ABC + ∠EDB = ∠ABC + ∠FED,整理得2(∠EDB - ∠FED)=∠ABC.

(3)画出图形如图2所示.∠FED、∠EDB与∠ABC之间的数量关系为2(∠FED + ∠EDB)-∠ABC = 360°.
解析:
∵EF//BC,
∴∠FEB = ∠ABC.又
∵∠BED = 180° - ∠EBD - ∠EDB = 180° - $\frac{1}{2}$∠ABC - ∠EDB,
∴∠FED = ∠FEB + ∠BED = ∠ABC + 180° - $\frac{1}{2}$∠ABC - ∠EDB = 180° + $\frac{1}{2}$∠ABC - ∠EDB,整理得2(∠FED + ∠EDB)-∠ABC = 360°.