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 sides)
AB = AC (Since, triangle is isosceles)
∠ADB = ∠ADC (90°)

Hence, by SAS axiom △ABD ~ △ACD. Since, triangles are similar so the ratio of the corresponding sides will be equal,

$\dfrac{BD}{DC} = \dfrac{AB}{AC} \\[1em] \dfrac{BD}{DC} = \dfrac{AC}{AC} \qquad [\because \text{AB = AC}] \\[1em] \dfrac{BD}{DC} = 1 \\[1em] BD = DC.$

Since, BD = DC so AD can be said as the perpendicular bisector of BC.

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