∠∠AOB = 90°∠DCE = 90°。

In the quadrilateral ODCE, OEC = 90°.

Namely CE⊥OB

∫C is on the bisector of∝∠ ∝∠AOB.

∴CD=CE (the distance from the point on the bisector of the angle is equal to both sides of the angle)

② Solution: Figure 2: The above conclusion is still valid. The reason for this is the following：

The intersection c is CK⊥OA, the vertical foot is k of CH⊥OB, and the vertical foot is H.

∵OM is the bisector of ∞∠AOB, CK⊥OA, CH⊥OB,

∴CK=CH,∠CKD=∠CHE=90，

∠≈ 1 and ∠2 are rotation angles,

∴∠ 1=∠2,

In △CKD and △CHE,

∠CKD=∠CHE

CK=CH

∠ 1=∠2

∴△CKD≌△CHE(ASA)，

∴CD=CE.

Figure 3: Just imitate the one in Figure 2. . . You should be able to!