1. 如图,在四边形ABCD中,AE平分∠BAD,DE平分∠ADC.
(1)若∠BAD=130°,∠CDA=140°,求∠EAD,∠EDA,∠AED的度数.
(2)若∠B+∠C=120°,求∠AED的度数.
(3)根据(2)的结论,请猜想∠B+∠C与∠AED之间的关系,并说明理由.

(1)
∵ ∠BAD=130°,∠CDA=140°,AE平分∠BAD,DE平分∠ADC,
∴ ∠EAD = 1/2 ∠BAD = 65°, ∠EDA = 1/2 ∠CDA = 70°,
∴ ∠AED = 180°−∠EAD −∠EDA = 180°−65°−70° = 45°.
(2)若∠B+∠C=120°,
∴ ∠BAD+∠CDA = 360°−120° = 240°.
∵ AE平分∠BAD, DE 平分∠ADC,
∴ ∠EAD + ∠EDA = 1/2(∠BAD+∠ADC) = 120°,
∴ ∠AED = 180°−120° = 60°.
(3)∠AED=1/2(∠B+∠C).理由:
∵ AE平分∠BAD,DE平分∠ADC,
∴ ∠EAD+∠EDA = 1/2 ∠BAD + 1/2 ∠ADC = 1/2(∠BAD+∠ADC) = 1/2(360°−∠B−∠C) = 180°−1/2(∠B+∠C),
∴ ∠AED = 180°−(∠EAD+∠EDA) = 1/2(∠B+∠C).

2. 如图,在△ABC中,∠ABC与∠ACB的平分线交于点D,DE⊥BC于点E.
(1)若∠BDE=60°,∠DCE=25°,求∠A的度数.
(2)若∠BDE−∠DCE=n°,解答以下问题:
①若n=30,求∠A的度数;
②试用含n的式子表示∠A,请说明理由.

(1)
∵ DE ⊥ BC,
∴ ∠BED = 90°,
∴ ∠DBE = 90°−∠BDE = 90°−60° = 30°.
∵ BD 平分∠ABC,CD 平分∠ACB,
∴ ∠ABC = 2∠DBE = 2×30° = 60°,∠ACB = 2∠DCE = 2×25° = 50°.在△ABC中,∠ABC = 60°,∠ACB = 50°,
∴ ∠A = 180°−∠ABC−∠ACB = 180°−60°−50° = 70°.
(2)①
∵ DE ⊥ BC,
∴ ∠BED = 90°,
∴ ∠DBE = 90°−∠BDE.
∵ BD平分∠ABC,CD 平分∠ACB,
∴ ∠ABC = 2∠DBE = 2(90°−∠BDE) = 180°−2∠BDE,∠ACB = 2∠DCE,在△ABC中,∠ABC = 180°−2∠BDE,∠ACB = 2∠DCE,
∴ ∠A = 180°−∠ABC−∠ACB = 180°−(180°−2∠BDE)−2∠DCE = 180°−180°+2∠BDE−2∠DCE = 2(∠BDE−∠DCE) = 2×30° = 60°.
②∠A=2n°.理由:
∵ DE ⊥ BC,
∴ ∠BED = 90°,
∴ ∠DBE = 90°−∠BDE.
∵ BD 平分∠ABC,CD 平分∠ACB,
∴ ∠ABC = 2∠DBE = 2(90°−∠BDE) = 180°−2∠BDE,∠ACB = 2∠DCE.在△ABC中,∠ABC = 180°−2∠BDE,∠ACB = 2∠DCE,
∴ ∠A = 180°−∠ABC−∠ACB = 180°−(180°−2∠BDE)−2∠DCE = 180°−180°+2∠BDE−2∠DCE = 2(∠BDE−∠DCE) = 2n°.