Given:
- Isosceles triangle ABC with AB = AC
- D is a point on BC

To prove:
AB^2 - AD^2 = BD × CD

Proof:

Step 1: Draw a diagram
- Draw an isosceles triangle ABC with AB = AC
- Draw a point D on BC

Step 2: Analyze the triangle
- Since AB = AC, triangle ABC is isosceles.
- The base angles, ∠ABC and ∠ACB, are equal.

Step 3: Identify important segments
- The segment AD is the altitude from vertex A to the base BC.
- The segments BD and CD are the distances from point D to the vertices B and C, respectively.

Step 4: Use the Pythagorean theorem
- In right triangle ABD, we have AB^2 = AD^2 + BD^2 (by the Pythagorean theorem).
- In right triangle ACD, we have AC^2 = AD^2 + CD^2 (by the Pythagorean theorem).
- Since AB = AC, we can rewrite the second equation as AB^2 = AD^2 + CD^2.

Step 5: Solve for AB^2 - AD^2
- Subtracting the equation from step 4 from the equation from step 3, we get:
AB^2 - AC^2 = (AD^2 + BD^2) - (AD^2 + CD^2)
Simplifying, we have:
AB^2 - AC^2 = BD^2 - CD^2

Step 6: Use the fact that AB = AC
- Since AB = AC, we can substitute AC^2 with AB^2 in the equation from step 5:
AB^2 - AB^2 = BD^2 - CD^2
Simplifying, we have:
0 = BD^2 - CD^2

Step 7: Rearrange the equation
- Adding CD^2 to both sides of the equation, we get:
CD^2 = BD^2
Rearranging, we have:
AB^2 - AD^2 = BD × CD

Conclusion:
We have proved that in an isosceles triangle ABC, AB^2 - AD^2 = BD × CD.