25. (8分)数学中,我们把有一个内角大于$180°$的四边形称为镖形.
(1)如图1,在镖形$ABCD$中,试探索内角$∠ A,∠ B,∠ D$与钝角$∠ BCD$之间的数量关系,并说明理由;
【拓展延伸】
(2)如图2,$∠ A+∠ B+∠ C+∠ D+∠ E=\_\_\_\_\_\_$;(用含$α$的代数式表示)
(3)如图3,已知直角$△ ABC$的直角顶点$A$落在直线$l$上,过点$B,C$分别作$l$的垂线段,垂足为$E$,$F$,若$∠ ABE,∠ ACF$的平分线交于点$D$,则$∠ D=\_\_\_\_\_\_$;
(4)如图4,在(3)的条件下,$BD_1,CD_1$分别为$∠ ABD,∠ ACD$的平分线,它们的交点为$D_1$;$BD_2,CD_2$分别为$∠ ABD_1,∠ ACD_1$的平分线,它们的交点为$D_2$;……以此类推,则$∠ D_{2025}=\_\_\_\_\_\_$.
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25. 【点拨】本题考查三角形外角的性质,平行线的判定与性质,能够熟练运用已有结论证明是解题的关键.
【解析】(1) $∠ BCD = ∠ A + ∠ B + ∠ D$. 理由如下:
如图,连接 $AC$,并延长至点 $E$,
$\therefore ∠ BCE = ∠ B + ∠ BAC$,$∠ DCE = ∠ D + ∠ DAC$,
$\therefore ∠ BCD = ∠ BCE + ∠ DCE = ∠ B + ∠ BAC + ∠ D + ∠ DAC = ∠ B + ∠ D + ∠ BAD$.
$\therefore$ 角的数量关系为 $∠ BCD = ∠ A + ∠ B + ∠ D$.

(2)由(1)得 $∠ BFC = ∠ A + ∠ B + ∠ C$,
$\therefore ∠ EFD = ∠ A + ∠ B + ∠ C$.
由(1)得 $∠ EOD = α = ∠ EFD + ∠ D + ∠ E$,
$\therefore ∠ A + ∠ B + ∠ C + ∠ D + ∠ E = α$. 故答案为 $α$.
(3)由(1)得 $∠ BAC = ∠ D + ∠ ABD + ∠ ACD = ∠ D + \dfrac{1}{2}(∠ ABE + ∠ ACF)$.
$\because BE⊥ l$,$CF⊥ l$,$\therefore BE// CF$,$\therefore ∠ EBC + ∠ BCF = 180°$,
$\therefore ∠ ABE + ∠ ABC + ∠ ACB + ∠ ACF = 180°$,
$\therefore ∠ ABE + ∠ ACF = 180° -90° =90°$,
$\therefore ∠ BAC = ∠ D + \dfrac{1}{2}×90°$,$\therefore ∠ D =90° -45° =45°$. 故答案为 $45°$.
(4)由(3)得 $90° = ∠ D + \dfrac{1}{2}×90° = ∠ D +45°$,
$90° = ∠ D_1 + \dfrac{1}{2}×45°$,
$90° = ∠ D_2 + (\dfrac{1}{2})^2×45°$,$90° = ∠ D_3 + (\dfrac{1}{2})^3×45°$,$\dots$
$\therefore ∠ D_{2025} + (\dfrac{1}{2})^{2025}×45° =90°$,$\therefore ∠ D_{2025} =90° - \dfrac{45°}{2^{2025}}$.
故答案为 $90° - \dfrac{45°}{2^{2025}}$.