Simone如图，已知$Rt\triangle ABC$中，$\angle ACB=90^{\circ}$，$CD\bot AB$于点$D$，$\angle BAC$的平分线分别交$BC$，$CD$于点$E$、$F$.
的有关信息介绍如下：
$\left(1\right)\because CD\bot AB$，$\therefore \angle CDB=90^{\circ}$，$\therefore \angle B+\angle BCD=90^{\circ}$，$\because \angle ACB=90^{\circ}$，$\therefore \angle ACD+\angle BCD=90^{\circ}$，$\therefore \angle ACD=\angle B$，$\because AE$平分$\angle BAC$，$\therefore \angle CAE=\angle BAE$，$\therefore \angle ACD+\angle CAE=\angle B+\angle BAE$，即$\angle CFE=\angle CEF$，$\therefore CF=CE$，即$\triangle CEF$是等腰三角形；$(2)AB=2AC$，理由是：$\because E$在线段$AB$的垂直平分线上，$\therefore AE=BE$，$\therefore \angle B=\angle BAE$，$\because \angle CAE=\angle BAE$，$\angle ACB=90^{\circ}$，$\therefore 3\angle B=90^{\circ}$，$\therefore \angle B=30^{\circ}$，$\therefore AB=2AC$；$(3)$方法一、过$E$作$EM\bot AB$于$M$，$\because AC=2.5$，$\angle ACB=90^{\circ}$，$\angle B=\angle CAE=30^{\circ}$，$\therefore AE=2CE$，设$CE=2$，则$AE=2x$，由勾股定理得：$AC^{2}+CE^{2}=AE^{2}$，即$2.5^{2}+x^{2}=\left(2x\right)^{2}$，解得：$x=\frac{5\sqrt{3}}{6}$，即$CE=\frac{5\sqrt{3}}{6}$，$\because AE$平分$\angle CAB$，$\angle ACB=90^{\circ}$，$EM\bot AB$，$\therefore EM=CE=\frac{5\sqrt{3}}{6}$，$\therefore \triangle ABE$的面积$S=\frac{1}{2}×AB×EM=\frac{1}{2}×5\times \frac{5\sqrt{3}}{6}=\frac{25\sqrt{3}}{12}$；方法二、由勾股定理得：$BC=2.5\sqrt{3}$，$\because CE=\frac{5\sqrt{3}}{6}$，$\therefore BE=BC-CE=\frac{5\sqrt{3}}{3}$，$\therefore \triangle ABE$的面积$S=\frac{1}{2}×BE×AC=\frac{1}{2}\times \frac{5\sqrt{3}}{3}\times 2.5=\frac{25\sqrt{3}}{12}$.
$\left(1\right)\because CD\bot AB$，$\therefore \angle CDB=90^{\circ}$，$\therefore \angle B+\angle BCD=90^{\circ}$，$\because \angle ACB=90^{\circ}$，$\therefore \angle ACD+\angle BCD=90^{\circ}$，$\therefore \angle ACD=\angle B$，$\because AE$平分$\angle BAC$，$\therefore \angle CAE=\angle BAE$，$\therefore \angle ACD+\angle CAE=\angle B+\angle BAE$，即$\angle CFE=\angle CEF$，$\therefore CF=CE$，即$\triangle CEF$是等腰三角形；$(2)AB=2AC$，理由是：$\because E$在线段$AB$的垂直平分线上，$\therefore AE=BE$，$\therefore \angle B=\angle BAE$，$\because \angle CAE=\angle BAE$，$\angle ACB=90^{\circ}$，$\therefore 3\angle B=90^{\circ}$，$\therefore \angle B=30^{\circ}$，$\therefore AB=2AC$；$(3)$方法一、过$E$作$EM\bot AB$于$M$，$\because AC=2.5$，$\angle ACB=90^{\circ}$，$\angle B=\angle CAE=30^{\circ}$，$\therefore AE=2CE$，设$CE=2$，则$AE=2x$，由勾股定理得：$AC^{2}+CE^{2}=AE^{2}$，即$2.5^{2}+x^{2}=\left(2x\right)^{2}$，解得：$x=\frac{5\sqrt{3}}{6}$，即$CE=\frac{5\sqrt{3}}{6}$，$\because AE$平分$\angle CAB$，$\angle ACB=90^{\circ}$，$EM\bot AB$，$\therefore EM=CE=\frac{5\sqrt{3}}{6}$，$\therefore \triangle ABE$的面积$S=\frac{1}{2}×AB×EM=\frac{1}{2}×5\times \frac{5\sqrt{3}}{6}=\frac{25\sqrt{3}}{12}$；方法二、由勾股定理得：$BC=2.5\sqrt{3}$，$\because CE=\frac{5\sqrt{3}}{6}$，$\therefore BE=BC-CE=\frac{5\sqrt{3}}{3}$，$\therefore \triangle ABE$的面积$S=\frac{1}{2}×BE×AC=\frac{1}{2}\times \frac{5\sqrt{3}}{3}\times 2.5=\frac{25\sqrt{3}}{12}$.
