Draw and describe the locus of vertices of all isosceles triangles having a common base.

△ABC is an isosceles triangle in which AB = AC.

From A, draw AD perpendicular to BC.

In △ABD and △ACD

AD = AD (Common side)
AB = AC (Since the, triangle is isosceles)
∠ADB = ∠ADC (90°)

Hence, by RHS congruence, △ABD ≅ △ACD.
Therefore, BD = DC

Since AD is perpendicular to BC and bisects BC, AD is the perpendicular bisector of BC.

Hence, the locus of vertices will be the perpendicular bisector of the base.