Construct an isosceles triangle whose base is $8cm$ and altitude $4cm$ and then another triangle whose sides are $1\frac{1}{2}$ times the corresponding sides of the isosceles triangle.

1. Draw a base $BC$ of length $8cm$.

2. Draw the perpendicular bisector of $BC$ and name it $PQ$. Let it intersect $BC$ in $M$.

3. From $M$, mark an arc $A$ at a distance of $4cm$ on $PQ$.

4. Join $B-A$ and $C-B$. $△ABC$ is the first isosceles triangle.

5. Extend $BC$ and $BA$. Place the pointed end of the compass at $C$ and mark an arc of radius $CM$ on the opposite side of $M$ on the extended segment $BC$. Mark the point as D (Since $CM$ is half of $BC$, $BD$ will be $BC+0.5BC=1.5BC$)

6. From $D$, draw a line parallel to $AC$ and let it intersect the extended segment $AB$ at $E$.

$△EBD$ is the required similar triangle.