Can a triangle be divided into four similar triangles such that not all four of the triangles are congruent to each other? I understand that you can divide a triangle into four congruent triangles by connecting the midpoints of each side. Can ANY NON-EQUILATERAL triangle be divided into four similar triangles with the restriction that not all four triangles can be congruent to each other?
As I explore this question, I keep running into dead ends, and I ask if any of you can help.
EDIT: You guys revealed that there are multiple ways to do this with right triangles. I've been experimenting with a general case and right triangles, but the closest I've gotten is splitting the triangle three times (on triangle ABC, drawing a line from Angle BAC that is perpendicular to Side BC, calling the point of intersection on Line BC Point D, then drawing lines from Angles ADB and ADC to be perpendicular with Lines AB and AC, respectively) yet I cannot prove that the triangles within ACD are similar to the triangles within ABD unless they are all within a right triangle. How to proceed?
 A: To obtain such a figure for a given triangle $\triangle ABC$ with angles $\alpha,\beta,\gamma$ where $|BC|\ne|CA|$ (and therefore $\alpha\ne\beta$),

*

*let $F$ be the intersection of the parallel to $AB$ through $C$ and the parallel to $BC$ through $A$ (so $AFBC$ is a parallelogram),

*construct line $\ell$ as tangent to the circumcircle of $ABC$ at $C$,

*let $D$ be the intersection of $\ell $ and $AC$,

*let $E$ be the intersection of $\ell$ and $BF$.


We have

*

*$\angle ABF = \angle BAC=\alpha$ (alternate angles as $FB\|AC$)

*$\angle FAB = \angle CBA=\beta$ (alternate angles as $FA\|BC$)

*$\angle BCE =\angle BAC=\alpha$ (inscribed angle theorem / chord-tangent theorem)

*$\angle CAD =\pi-\angle FAC=\pi-(\alpha+\beta)=\gamma$ (supplementary angle and angle sum in triangle)

*$\angle EBC=\pi-\angle CBF=\pi-(\alpha+\beta)=\gamma$ (supplementary angle and angle sum in triangle)

*$\angle DCA = \pi-\angle ACE=\pi-(\alpha+\gamma)=\beta$ (supplementary angle and angle sum in triangle)

*$\angle ADC=\alpha$ (angle sum in triangle)

*$\angle CEB=\beta$ (angle sum in triangle)

*$\angle BFA=\gamma$ (angle sum in triangle)

Thus triangles $ABC$, $DCA$, $CEB$, $BAF$, $DEF$ are all similar.
But they are not all congruent: By comparing the sides opposing angle $\alpha$, we find
$$ {\triangle DCA}:{\triangle ABC}=|CA|:|BC|\ne 1:1$$
Finally, in order to partition $\triangle ABC$ instead of extending it, we need only perform a similarity transformation that maps $\triangle DEF$ to $\triangle ABC$.
A: The drawing explains how you do it.
There are at least three different solutions.
You did not ask for a proof so I just demonstrated a potential solution

A: For a right isosceles triangle, bisect its right angle.  Select one of the smaller triangles thus formed and bisect that right angle.  Do the latter step again.  QEF (Latin, which was to be done).
