Dilation, shrink a triangle by $30\%$ I would like to know how I shrink a triangle by $30\%$ using dilation not changing its center?
 A: Multiply every coordinate of the original triangle by the scale factor $\left(\frac 7{10}\right)$
Also see this for  reference.
A: Analytically: translate the center to the origin, shrink everything, translate back.
Geometrically: move the points on each line through the center toward the center by decreasing length by a factor $0.7$.
This changes all lengths to 70% of what they were. It will fix whatever point you have decided is the "center" - probably the intersection of the medians. For areas, use factor $\sqrt{0.7}$. 
A: I presume you want to shrink the Area by $30\%$.
Assume the center is at $p$. Then you start by translating the triangle so that its center is in $0$, and then note that the area scales like the square of the scaling factor, so you have to scale the coordinates by $\sqrt{0.7}$, and then translate it back to its original position.
Composing these maps gives:
$$f(\vec{x})=\sqrt{0.7}\ (\vec{x}-\vec{p})+\vec{p}$$
A: Affine solution:

The task is to shrink the triangle $ABC$ "around the center" $O$.
To create a dilation trisect $BC$. Let the length of the red segment be $\frac23$ $BC$. 
Then from $O$ draw lines through the end points of the red segment. Draw a line (white)  through $C$ so that it is parallel to $OB$...
So, you can construct the thick black segment which is parallel to the red segment and its length is the same as that of $BC$.
The red segment and the thick black segment determine a dilation with center $O$.
The yellow lines show how to construct the image of the thick black point on $BC.$  Moving this black point, the resulting images will trace the shrunk image of $BC$. Actually, you don't have to trace. It is enough to construct one point and then to construct a parallel with $BC$. 
Do the same with the other sides of the triangle. Then the vertices will shop up as well.
