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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?

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    You can do this (in multiple ways) with right triangles. – Blue Nov 01 '20 at 00:02
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    Any isosceles triangle admit such a decomposition. Aside from the special case of a $1 : 1 : \sqrt{2}$ isosceles triangle, you can cut any isosceles triangle into two congruent right triangle and then cut each right triangle into two similar but not congruent triangles. For the special case $1: 1 : \sqrt{2}$, you need another cutting procedure but it can be done. – achille hui Nov 01 '20 at 00:35
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    Related: "The Similar Triangles Point". This illustrates the three ways to dissect a scalene triangle as described (where we allow two sub-triangles to be congruent). The precise dissection isn't given, but can be readily deduced. (The point of that question is interesting in itself.) – Blue Nov 01 '20 at 01:16

3 Answers3

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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 $BF$.

enter image description here

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 $\triangle ABC$.

Hagen von Eitzen
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  • "Let F bethe intersection of the parallel to AB through C" -- there must have been some weird drug in the omelette I had this morning because I am seeing the parallel to AC through B instead. May want to check typing. – Oscar Lanzi Nov 01 '20 at 16:56
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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 enter image description here

Moti
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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).

Oscar Lanzi
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