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Perpendicular bisector of Mathematical Line Segment
Solution: (1) Let ∠ cab = x, ∠ CBA = y.

∠∠ACB = 90°

∴x+y=90

∵AF divides CD vertically, BG divides CE vertically.

∴AC=AD,BC=BE

∴△acd:∠ACD =∠ADC =( 180-∠cab)/2 = 90- 1/2 x

△BCE:∠BCE =∠BEC =( 180-∠CBA)/2 = 90- 1/2y

∠△DCE:∠ECD = 180-∠ADC-∠BEC

∴∠ECD = 180-(90- 1/2 x)-(90- 1/2y)

= 1/2 (x+y)

= 1/2 x90

=45

(2)∠ECD = 90- 1/2a; The reason is:

Let ∠ cab = x, ∠ CBA = y.

∫∠ACB = a

∴x+y= 180 -a

∵AF divides CD vertically, BG divides CE vertically.

∴AC=AD,BC=BE

∴△acd:∠ACD =∠ADC =( 180-∠cab)/2 = 90- 1/2 x

△BCE:∠BCE =∠BEC =( 180-∠CBA)/2 = 90- 1/2y

∠△DCE:∠ECD = 180-∠ADC-∠BEC

∴∠ECD = 180-(90- 1/2 x)-(90- 1/2y)

= 1/2 (x+y)

= 1/2 x( 180 -a)

=90 - 1/2 a

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