1. 🎯|跨学科融合（2025·连云港期末）【习题再现】(1)苏科版初中数学教材七上有一题:如图①,AB//CD,点E在AB,CD之间.写出∠AEC,∠A,∠C之间的数量关系,并说明理由；
【迁移思考】(2)小明在完成这题的探究后,又进一步变式思考:
①如图②,在长方体盒子底部有一面平面镜,点A处有一个光源,光线的入射角等于反射角,法线OE与平面镜l垂直,即OE⊥BC,垂足为O,入射光线经过镜面反射后,恰好经过点D.小明认为∠AOB=∠COD,请帮小明说明理由；
②如图③,在长方体盒子里放置4块平面镜AB,BC,CD,DA,其中AD//CB,若光线从AD上的E处射出,在平面镜AB上经点F反射后,到达BC上的点G……其传播路径为E→F→G→H→E→F……请判断∠EFG与∠GHE的数量关系,并说明理由.

(1)∠AEC=∠A+∠C,理由如下:过点E作EM//AB,如图所示:
因为AB//CD,所以AB//EM//CD,所以∠AEM = ∠A,∠CEM=∠C,所以∠AEM+∠CEM=∠A+∠C.因为∠AEC=∠AEM+∠CEM,所以∠AEC=∠A+∠C.
(2)①理由如下:因为OE⊥BC,所以∠EOC=∠EOB=90°,所以∠AOE+∠AOB = ∠DOE+∠COD = 90°.因为光线的入射角等于反射角,所以∠AOE = ∠DOE,所以∠AOB = ∠COD.
②∠EFG = ∠GHE,理由如下:由①的结论得∠AEF = ∠DEH,∠BGF = ∠CGH,所以∠AEF + ∠BGF = ∠DEH + ∠CGH.因为AD//CB,由(1)的结论得∠EFG = ∠AEF + ∠BGF,∠GHE = ∠DEH+∠CGH,所以∠EFG=∠GHE.

2. 如图,已知点 P 在 AB,CD 两线之间,且在 BC 所在直线的左侧.
(1)如图①,当$AB// CD,∠BPC=α$时,
①若 BO 平分$∠ABP$,CO 平分$∠DCP$,则$∠BOC=$;
②若$∠ABO=\frac {1}{3}∠ABP,∠DCO=\frac {1}{3}∠DCP$,则$∠BOC=$;
③若$∠ABO=\frac {1}{n}∠ABP,∠DCO=\frac {1}{n}∠DCP$,则$∠BOC=$.
(2)如图②,当 AB 与 CD 相交,点 A,D 重合时,猜想$∠BPC,∠B,∠C$与$∠A$之间的数量关系,并说明理由.
(3)如图③,直接运用(2)的结论探究下列问题:
①若 BO 平分$∠ABP$,CO 平分$∠ACP$,当$∠BPC=120^{\circ },∠BOC=95^{\circ }$时,求$∠A$的度数;
②若$∠ABO=\frac {1}{n}∠ABP,∠ACO=\frac {1}{n}∠ACP$,当$∠BPC=α,∠BOC=β$时,求$∠A$的度数.

＞＞ 进一步挑战进阶专题:P143 专题24~P147 专题27

(1)①$\frac{α}{2}$ ②$\frac{α}{3}$ ③$\frac{α}{n}$
【解析】①如图①,分别过点O,P作OE//AB,PQ//AB,因为AB//CD,所以OE//PQ//AB//CD,所以∠ABP = ∠BPQ,∠DCP = ∠CPQ,∠ABO = ∠BOE,∠DCO = ∠COE,所以∠ABP+∠DCP = ∠BPC,∠ABO+∠DCO = ∠BOC.因为∠BPC = α,BO平分∠ABP,CO平分∠DCP,所以∠ABO = ∠PBO = $\frac{1}{2}$∠ABP,∠DCO = ∠PCO = $\frac{1}{2}$∠DCP,所以∠BOC = $\frac{1}{2}$∠ABP + $\frac{1}{2}$∠DCP = $\frac{1}{2}$(∠ABP+∠DCP) = $\frac{1}{2}$∠BPC = $\frac{α}{2}$.
②由①得∠ABP+∠DCP = ∠BPC,∠ABO+∠DCO = ∠BOC,因为∠ABO = $\frac{1}{3}$∠ABP,∠DCO = $\frac{1}{3}$∠DCP,所以∠BOC = $\frac{1}{3}$∠ABP+$\frac{1}{3}$∠DCP = $\frac{1}{3}$(∠ABP+∠DCP) = $\frac{1}{3}$∠BPC = $\frac{α}{3}$.
③由①得∠ABP+∠DCP = ∠BPC,∠ABO+∠DCO = ∠BOC,因为∠ABO = $\frac{1}{n}$∠ABP,∠DCO = $\frac{1}{n}$∠DCP,所以∠BOC = $\frac{1}{n}$∠ABP+$\frac{1}{n}$∠DCP = $\frac{1}{n}$(∠ABP+∠DCP) = $\frac{1}{n}$∠BPC = $\frac{α}{n}$.

(2)∠A+∠B+∠C = ∠BPC,理由如下:如图②,作射线BF,分别过点P,A,C作PQ//BF,AG//BF,CE//BF,则PQ//BF//AG//CE,所以∠ABF = ∠BAG,∠ACE = ∠CAG,∠PBF = ∠BPQ,∠PCE = ∠CPQ,所以∠BPC = ∠PBF + ∠PCE.因为∠PBF = ∠ABF + ∠ABP,∠PCE = ∠ACE + ∠ACP,所以∠BPC = ∠ABF + ∠ABP + ∠ACE + ∠ACP = ∠BAC+∠ABP+∠ACP,即原图中∠A+∠B+∠C = ∠BPC.
(3)①由(2)可得∠A+∠ABP+∠ACP = ∠BPC = 120°,∠A+∠ABO+∠ACO = ∠BOC = 95°,因为BO平分∠ABP,CO平分∠ACP,所以∠ABO = ∠OBP = $\frac{1}{2}$∠ABP,∠ACO = ∠OCP = $\frac{1}{2}$∠ACP,所以∠A + $\frac{1}{2}$(∠ABP + ∠ACP) = ∠BOC = 95°.因为∠A + ∠ABP + ∠ACP - (∠A + ∠ABO + ∠ACO) = ∠BPC - ∠BOC = 25°,即$\frac{1}{2}$(∠ABP + ∠ACP) = 25°,所以∠ABP+∠ACP = 50°,所以∠A = ∠BPC - (∠ABP+∠ACP) = 70°.
②因为∠ABO = $\frac{1}{n}$∠ABP,∠ACO = $\frac{1}{n}$∠ACP,所以∠A + ∠ABP+∠ACP = ∠BPC = α,∠A+∠ABO+∠ACO = ∠BOC = β,所以∠A + $\frac{1}{n}$(∠ABP + ∠ACP) = ∠BOC = β,由①得∠BPC-∠BOC = α-β,所以$\frac{n-1}{n}$(∠ABP+∠ACP) = α-β,即∠ABP+∠ACP = $\frac{n}{n-1}$(α-β),所以∠A = ∠BPC-(∠ABP+∠ACP) = α-$\frac{n}{n-1}$(α-β) = $\frac{n}{n-1}β - \frac{1}{n-1}α$.