Proof that in an isosceles triangle, the altitude from the vertex bisects the base:

Given: an isosceles triangle ABC, where AB = AC.

To prove: the altitude from the vertex A bisects the base BC.

Proof:

1. Draw an isosceles triangle ABC with AB = AC.
2. Draw the altitude from the vertex A, intersecting the base BC at point D.

Proof by contradiction:

Assume that the altitude AD does not bisect the base BC.

Case 1: AD is not perpendicular to BC.
In this case, AD and BC would intersect at a point E, such that AE ≠ DE.

3. Draw line segments AE and DE.
4. Since AD is the altitude, it is perpendicular to BC. Therefore, ∠ADE and ∠AED are right angles.
5. In triangle ADE, AD = AD (common side) and ∠ADE = ∠AED (both are right angles).
6. By the Side-Angle-Side (SAS) congruence criterion, triangle ADE is congruent to itself.
7. Therefore, AE = DE (corresponding parts of congruent triangles are equal).
8. But this contradicts our assumption that AE ≠ DE.
9. Hence, our assumption that AD is not perpendicular to BC is false.

Case 2: AD is perpendicular to BC, but does not bisect BC.
In this case, AD divides BC into two segments, BD and DC, such that BD ≠ DC.

10. Draw line segments BD and DC.
11. By the definition of an altitude, AD is perpendicular to BC. Therefore, ∠ADB and ∠ADC are right angles.
12. In triangle ADB and ADC, AD = AD (common side), ∠ADB = ∠ADC (both are right angles), and AB = AC (given isosceles triangle).
13. By the Side-Angle-Side (SAS) congruence criterion, triangles ADB and ADC are congruent.
14. Therefore, BD = DC (corresponding parts of congruent triangles are equal).
15. But this contradicts our assumption that BD ≠ DC.
16. Hence, our assumption that AD does not bisect BC is false.

Conclusion: In both cases, our assumptions led to contradictions. Therefore, the altitude from the vertex A must bisect the base BC in an isosceles triangle.