9. 如圖,$B$,$C$,$D$是不在同一直線上的三點(diǎn),且$∠CDE + ∠BCD - ∠ABC = 180^{\circ}$。
(1)如圖①,試說明：$AB// DE$。
(2)$DG平分∠EDC$,點(diǎn)$P是DG$上一點(diǎn),過點(diǎn)$P作射線PB$,設(shè)$∠1 = \alpha$。
①如圖②,若$PD// BC$,$∠ABC = 2∠3$,求$∠C$的度數(shù)；(用含$\alpha$的式子表示)
②如圖③,若$∠3 + \frac{1}{2}∠C = 90^{\circ}$,判斷$∠1與∠2$的數(shù)量關(guān)系,并說明理由。
答案：9.(1)如圖①,延長DC交AB于點(diǎn)F,因?yàn)椤螧CD + ∠BCF = 180°,∠BCF + ∠BFC + ∠FBC = 180°,所以∠BCD = ∠BFC + ∠FBC,所以∠BFC = ∠BCD - ∠ABC.因?yàn)椤螩DE + ∠BCD - ∠ABC = 180°,即∠CDE + ∠BFC = 180°,所以AB//DE(同旁內(nèi)角互補(bǔ),兩直線平行).
(2)①因?yàn)镻D//BC,所以∠2 = ∠3.又因?yàn)椤螦BC = 2∠3,∠ABC = ∠1 + ∠2,∠1 = α,所以∠1 = ∠2 = ∠3 = α,∠ABC = 2α.因?yàn)镈G平分∠EDC,∠CDG + ∠C = 180°,所以 $\frac{1}{2}$∠EDC + ∠C = 180°,所以∠EDC = 360° - 2∠C.因?yàn)椤螮DC + ∠BCD - ∠ABC = 180°,即∠EDC = 180° - ∠C + 2α,所以360° - 2∠C = 180° - ∠C + 2α,即∠C = 180° - 2α.
②∠1 = ∠2.理由如下:
如圖②,過點(diǎn)C作CH平分∠BCD交DG于點(diǎn)H,則∠4 = ∠5 = $\frac{1}{2}$∠BCD.
因?yàn)镈G平分∠EDC,所以∠CDG = $\frac{1}{2}$∠EDC,所以∠5 + ∠CDG = $\frac{1}{2}$(∠BCD + ∠EDC),又因?yàn)椤螩DE + ∠BCD - ∠ABC = 180°,所以∠5 + ∠CDG = $\frac{1}{2}$(180° + ∠ABC) = 90° + $\frac{1}{2}$∠ABC,所以∠PHC = 90° + $\frac{1}{2}$∠ABC.因?yàn)椤? + $\frac{1}{2}$∠BCD = 90°,所以∠3 + ∠4 = 90°,所以∠PBC = 360° - ∠3 - ∠4 - ∠PHC = 360° - 90° - 90° - $\frac{1}{2}$∠ABC = 180° - $\frac{1}{2}$∠ABC.又因?yàn)椤螾BC + ∠2 = 180°,所以∠2 = $\frac{1}{2}$∠ABC,所以∠1 = ∠2.
