25.如图,在四边形ABCD中,AB//DC,对角线AC,BD交于点O,OB=OD,且DB平分∠ADC,点E为AB边的中点,连接OE,连接CE交DB于点F.
(1)求证:四边形ABCD是菱形;
(2)若∠AOE=28°,∠CEB=38°,求∠CFB的度数.

25.(1)证明:$\because AB // CD, \therefore ∠ ABO = ∠ CDO$,
在 $△ ABO$ 和 $△ CDO$ 中, $\begin{cases} ∠ ABO = ∠ CDO, \\ OB = OD, \\ ∠ AOB = ∠ COD, \end{cases}$
$\therefore △ ABO ≌ △ CDO(\mathrm{ASA}), \therefore AB = CD, \therefore$ 四边形 $ABCD$ 是平行四边形. $\because DB$ 平分 $∠ ADC$,
$\therefore ∠ ADB = ∠ CDO, \therefore ∠ ABO = ∠ ADB$,
$\therefore AD = AB, \therefore$ 平行四边形 $ABCD$ 是菱形.
(2)解:由(1)可知,四边形 $ABCD$ 是菱形,
$\therefore AC ⊥ BD, \therefore ∠ AOB = 90°$.
$\because$ 点 $E$ 为 $AB$ 边的中点, $\therefore OE = \frac{1}{2}AB = AE$,
$\therefore ∠ OAE = ∠ AOE = 28°$,
$\therefore ∠ ABO = 90° - 28° = 62°, \therefore ∠ CFB = ∠ CEB + ∠ ABO = 38° + 62° = 100°$, 即 $∠ CFB$ 的度数为 $100°$.

26.如图，直线AB分别与y轴、x轴交于点A(0,3√3),B(3,0),点C的坐标为(-3,0),D为直线AB上一动点，连接CD交y轴于点E.
(1)求直线AB的解析式；
(2)若$S_{△ COE}=S_{△ ADE}$，求点D的坐标.

26.解:(1)设直线 $AB$ 的解析式为 $y = kx + b$.
由题意得 $\begin{cases} b = 3\sqrt{3}, \\ 3k + b = 0, \end{cases}$ 解得 $\begin{cases} k = -\sqrt{3}, \\ b = 3\sqrt{3}, \end{cases}$
$\therefore$ 直线 $AB$ 的解析式 $y = -\sqrt{3}x + 3\sqrt{3}$.
(2)$\because S_{△ COE} = S_{△ ADE}, \therefore S_{△ AOB} = S_{△ CBD}$, 即 $\frac{1}{2} × 6 · y_D = \frac{1}{2} × 3 × 3\sqrt{3}, \therefore y_D = \frac{3\sqrt{3}}{2}$. 当 $y = \frac{3\sqrt{3}}{2}$ 时, 有 $-\sqrt{3}x + 3\sqrt{3} = \frac{3\sqrt{3}}{2}$, 解得 $x = \frac{3}{2}, \therefore$ 点 $D$ 的坐标为 $(\frac{3}{2}, \frac{3\sqrt{3}}{2})$.