How can I get 17 VDC from 25 VDC with current may up to 10 A?
25V AC means 25V effective, which is sqrt(2)*25V = 35V 0-to-top. Hence when you rectify this to DC you can get up to 35V, because the capacitor will be filled according to the top of the AC wave.

1N4148 diodes are MUCH to feeble for this application. Use at least rated for 10A plus ample margin. You might need to limit the inrush current when you first plug your power supply in.

You can indeed reduce voltage with a resistor divider, but this is only practical when the current through the resistors is much smaller than the load current, which is clearly not the case for you.

A better way is to use a voltage regulator chip, like the 780x series or an LM317. These chips behave a s series resistors, but they constantly adapt their resistance to achieve the desired output voltage. But like any resistor, they will dissipate heat according to the formula I * V = P, which is your case is 10 * (35 - 17) = 180W. That is a lot of heat, which makes this solution impractical.

The best way to reduce (or increase!) a voltage at a substantial current is to use a switching regulator. Designing such a thing for the voltages and especially the current you need is difficult, I guess it is way above your head.

Luckily, 17V adapters are available in most computer shops, because these values are often used by laptops. 10A is a bit heavy, but they might be available. I suggest you buy one :)