If two triangles $ABC$ and $DBC$ have a common base $BC$, and $AD$ is parallel to $BC$, and $△ABC$ is an isosceles triangle, then the perimeter of $△ABC$ is less than that of $△DBC$.

If we extend $BA$ and take a point $C’$ such that $AC=AC’$, then

$$AB+AC=AB+AC’=BC’. \qquad [1]$$

Since $AD$ is the perpendicular bisector of $CC’$,

$$DC=C’ D. \qquad [2]$$

For $△BC’ D$,

$$BC'<C’ D+DB,$$

$∴ \ AB+AC<DC+DB,$          (from $[1]$ and $[2]$)

$$∴ \ AB+AC+BC<DB+DC+BC.$$

In other words, the perimeter of $△ABC$ is smaller than that of $△DBC$.

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