《2025年点石成金金牌夺冠八年级数学上册人教版辽宁专版》

3. 如图，点$B$，$C$分别在射线$AM$，$AN$上，点$E$，$F$在$\angle MAN$内部的射线$AD$上，已知$AB = AC$，$\angle BED = \angle BAC = \angle CFD$.
(1)求证：$\triangle ABE \cong \triangle CAF$.
(2)试判断$BE$，$CF$，$EF$之间的数量关系，并说明理由.

答案： 3.
(1)证明：
∵∠BED = ∠BAE + ∠ABE，∠BAC = ∠BAE + ∠CAF，∠CFD = ∠ACF + ∠CAF，且∠BED = ∠BAC = ∠CFD，
∴∠ABE = ∠CAF，∠BAE = ∠ACF.
在△ABE和△CAF中，$\begin{cases} ∠ABE = ∠CAF, \\ AB = CA, \\ ∠BAE = ∠ACF \end{cases}$
∴△ABE≌△CAF(ASA).
(2)BE = CF + EF.
理由：由
(1)得，△ABE≌△CAF.
∴AE = CF，BE = AF.
∴BE = AF = AE + EF = CF + EF.

4. 已知$\triangle ABC$是等边三角形，点$D$为射线$AC$上一点，连接$BD$，$\angle DBE = 120^{\circ}$，$BD = BE$.
(1)如图1，过点$E$作$EF // AC$交边$AB$于点$F$，求证：$FB = CD$.
(2)如图2，当点$D$在边$AC$上时，连接$CE$交边$AB$于点$G$. 若$AB = 4$，$BG = 1$，求证：$BD \perp AC$.

答案：
4.
(1)证明：
∵△ABC是等边三角形，
∴∠ABC = ∠ACB = ∠A = 60°.
∴∠BCD = 180° - ∠ACB = 120°.
∵EF//AC，
∴∠EFA = ∠A = 60°.
∴∠BFE = 180° - ∠EFA = 120°.
∴∠BFE = ∠BCD.
∵∠DBE = 120°，
∴∠EBF + ∠DBC = ∠DBE - ∠ABC = 60°.
∵∠ACB = ∠DBC + ∠D = 60°，
∴∠EBF = ∠D.
在△EFB和△BCD中，$\begin{cases} ∠EFB = ∠BCD, \\ ∠EBF = ∠D, \\ EB = BD \end{cases}$
∴△EFB≌△BCD(AAS).
∴FB = CD.
(2)证明：如图，过点E作EF//AC交AB的延长线于点F，
∴∠F = ∠A.
∵△ABC是等边三角形，
∴∠ABC = ∠ACB = ∠A = ∠F = 60°，AB = BC = AC.
∴∠FBC = 180° - ∠ABC = 120°.
∵∠DBE = 120°，
∴∠CBD + ∠EBF = 360° - ∠FBC - ∠DBE = 120°.
∵∠BDC + ∠CBD = 180° - ∠ACB = 120°，
∴∠EBF = ∠BDC.
在△EFB和△BCD中，$\begin{cases} ∠F = ∠BCD, \\ ∠EBF = ∠BDC, \\ EB = BD \end{cases}$
∴△EFB≌△BCD(AAS).
∴FB = CD，EF = BC = AC.
在△EGF和△CGA中，$\begin{cases} ∠EGF = ∠CGA, \\ ∠F = ∠A, \\ EF = CA \end{cases}$
∴∠EGF≌△CGA(AAS).
∵AB = 4，BG = 1，
∴GF = GA = AB - BG = 3.
∴CD = FB = GF - BG = 3 - 1 = 2.
∴AD = AC - CD = AB - CD = 2.
∴CD = AD.
∵BC = BA，
∴BD⊥AC.