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For any acute and right triangle, two of the intersections(C and B) of the two inscribed squares and the vertex A of the triangle are collinear.

I have a proof for the theorem, but I have not found any specific name for it. I searched "inscribed squares in a triangle," and I still could not find the same shape. Can anyone provide a specific name for this, or a way that I can find the theorem online?

For the similar question concerning the cubes inscribed in a tetrahedron, please visit Extend squares inscribed in a triangle to the cubes inscribed in a tetrahedron .

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$\tan(\angle ACB) = \frac{IK}{KC}$

$\because LG = IK$

$\therefore \frac{LG}{KC} = \frac{IK}{KC} = \tan(\angle ACB)$


$\frac{LH}{JC} = \tan(\angle ACB)$

$\because \angle LHO = \angle LGO = \angle OLC = \angle OKC = 90^{\circ}$

$\therefore \angle JCK = \angle GLH = 360^{\circ} - \angle LHO - \angle LGO - \angle GOH = 360^{\circ} - \angle OJC - \angle OKC - \angle JOK $

$\because \frac{LH}{JC} = \frac{LG}{KC} = \tan(\angle ACB), \angle JCK = \angle GLH$

$\therefore \Delta LHG \simeq \Delta CJK $ (SAS)

$\therefore \angle LGH = \angle JKC$

$\because \angle LGO = \angle OKC = 90^{\circ}$

$\therefore \angle HGO = \angle OKJ = \angle LGO - \angle LGH = \angle OKC -\angle JKC$

Similarly, $\angle GHO = \angle OJK$

$\therefore \Delta GOH \simeq \Delta KOJ$ (AA)

$\therefore \frac{GH}{JK} = \frac{LG}{KC}$

$\therefore \frac{GO}{OK} = \frac{LG}{KC} = \frac{GH}{JK}$

$\because \angle LGO = \angle OKC$

$\therefore \Delta LGO \simeq \Delta CKO$ (SAS)

$\therefore \angle GOL = \angle KOC$

Similarly, $\angle HOL = \angle JOC$

$\because \angle IOJ = \angle POK$

$\therefore \angle LOG + \angle IOJ + \angle JOC = \angle COK + \angle POK + \angle HOL = \frac{1}{2}(360^{\circ}) = 180^{\circ}$

$\therefore$ Points C, O, L are collinear


  • 1
    $\begingroup$ Which angle is 90 degrees? (unclear in diagram) $\endgroup$
    – coffeemath
    Aug 28, 2018 at 1:25
  • 2
    $\begingroup$ I added a new picture. Is that what you want me to make clear about? $\endgroup$
    – Larry
    Aug 28, 2018 at 1:29
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    $\begingroup$ I've never heard of this before. It may be genuinely new. $\endgroup$
    – mweiss
    Aug 28, 2018 at 2:59
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    $\begingroup$ I don't know if the very-interesting theorem has a name ("Larry's Theorem"?), but I know it can be generalized a bit: If you broaden the definition of "inscribed square" to "a square with vertices on the extended sides of a triangle" (in particular, with two adjacent vertices on a particular side-line), then your result applies to obtuse triangles, too. Moreover, under this broader notion, there are two "inscribed" squares on each side-line; a "small" and a "large". Your result applies to the large squares, as well. (Mixing large and small doesn't seem obviously fruitful, but who knows?) $\endgroup$
    – Blue
    Aug 28, 2018 at 11:06
  • 3
    $\begingroup$ @EdPegg: Sadly, the three lines aren't concurrent. $\endgroup$
    – Blue
    Aug 30, 2018 at 4:19

3 Answers 3


In light of the Bailey and Detemple (B&T) article mentioned in @Larry's answer, I'm updating my response. See the Edit History for a previous version of this answer.

As B&T describe, the key notion here is a general property of squares inscribed in angles; having a square's vertex lie on the third side of a triangle is an unnecessary restriction. (That restriction would have some relevance if the three resulting lines of intersection for a triangle were concurrent; alas, they are not.) But in fact, the property generalizes beyond squares to similar parallelograms.

Visually, the result is clear:

enter image description here

Articulating it in words is a little tricky. Bear with me.

Definitions. Given an angle with side-lines $a$ and $b$, we call a parallelogram "inscribed with base-line $a$" if two adjacent vertices (the "base vertices") lie on $a$ and a third vertex lies on $b$. We'll call the side-lines not parallel to $a$ the "lateral lines" of the parallelogram.

Let an angle have vertex $O$ and side-lines $a$ and $b$. Inscribe a parallelogram with base vertices $A$ and $A^\prime$ on $a$, and with $A^{\prime\prime}$ (adjacent to $A$) on $b$; let $A^\star$ be the fourth vertex. Likewise, inscribe parallelogram $\square B^\prime B B^{\prime\prime} B^\star$ with base-line $b$.

Theorem. If the parallelograms are similar and "compatibly slanted" (that is, if $\square A^\prime A A^{\prime\prime}A^\star \sim \square B^\prime B B^{\prime\prime} B^{\star}$, and $\angle OAA^{\prime\prime}\cong\angle OBB^{\prime\prime}$), then $O$ is collinear with the points where the lateral lines through $A$ and $A^\prime$ meet the respective lateral lines through $B$ and $B^\prime$.

For proof, let the lateral lines through $A$ and $B$ meet at $P$. Let $\overleftrightarrow{OP}$ meet the lateral line through $A^\prime$ at $P^\prime$. Via the similarity $\triangle OAP \sim OA^\prime P^\prime$, we deduce $$\frac{|PP^\prime|}{|OP|} = \frac{|AA^\prime|}{|OA|} = \frac{|AA^\prime|}{|AA^{\prime\prime}|}\frac{|AA^{\prime\prime}|}{|OA|} \tag{1}$$ The parallelogram similarity allows us to re-write the first factor in $(1)$, while $\triangle OAA^{\prime\prime}\sim\triangle OBB^{\prime\prime}$ allows us to re-write the second: $$\frac{|PP^\prime|}{|OP|} = \frac{|BB^\prime|}{|BB^{\prime\prime}|}\frac{|BB^{\prime\prime}|}{|OB|} \tag{2}$$ We may conclude, then, that $P^\prime$ lies on the lateral line through $B^\prime$, giving the Theorem. $\square$

Note that the Theorem holds for arbitrary pairs of parallelograms inscribed in an angle; moreover, it holds whether $\overrightarrow{PP^\prime}$ points in the same or opposite direction as $\overrightarrow{OP}$. @Larry and B&T restrict the parallelograms to squares, and @Larry constrains the "fourth vertices" $A^\star$ and $B^\star$ to a third line (B&T consider that as a special case). They all also implicitly assume the "same direction" configuration, but here we see that the "opposite direction" case is equally valid:

enter image description here

Here's a related (new?) result, for which I currently have only an ugly coordinate proof.

A triangle admits three more "inscribed" squares; these have opposite vertices on a particular side-line, and one vertex on each other side-line. Given $\triangle ABC$, draw the new squares associated with sides $\overline{AB}$ and $\overline{CA}$. The side-lines of the squares meet in four point-pairs, each of which are collinear with vertex $A$. (The lines also meet in another eight points that seem to have no remarkable collinearity property.)

In the figure, the four green lines show collinearity of top triangle vertex with four pairs of points-of-intersection of two squares' side-lines. When two lines meet at one of these points, the lines parallel to those lines meet at the other point of the pair.

enter image description here

This result, too, generalizes to parallelograms that need not have "fourth vertices" constrained to a third line. Here's an image:

enter image description here

  • 1
    $\begingroup$ Now I understand what you meant by the "large" square. That is amazing! $\endgroup$
    – Larry
    Sep 2, 2018 at 13:34
  • $\begingroup$ Nice proof, and I think it'd be even easier to understand by removing the length labels from the first diagram. $\endgroup$
    – user210229
    Sep 9, 2018 at 17:54
  • $\begingroup$ @Larry: Thanks for the bounty! :) BTW: You should post the tetrahedron variant as a separate question. $\endgroup$
    – Blue
    Sep 12, 2018 at 11:14
  • $\begingroup$ :) You’re welcome. I was thinking about posting it as a separate question, but I thought the question would be easier to understand with the context. Maybe I should post another question. $\endgroup$
    – Larry
    Sep 12, 2018 at 11:57
  • 1
    $\begingroup$ @Larry: Post another question, but add a link back to this one for context. (And add a link from this one to the other for people who might be interested.) Simple! :) $\endgroup$
    – Blue
    Sep 12, 2018 at 12:42

Here is an alternative proof, using the diagram below.

As an aside, note that proving $Q'Q \parallel R'P$ would allow us to conclude the rest from a special case of Desargues's theorem. We would have:

  • $Q'S \parallel R'O$, $\ SQ \parallel OP$, $\ QQ' \parallel PR'$, so
  • $Q'S \cap R'O$ and $SQ \cap OP$ and $QQ' \cap PR'$ are all collinear on the line at infinity, so
  • the triangles $Q'SQ$ and $R'OP$ are in perspective axially, so
  • the triangles are in perspective centrally, and
  • $Q'R'$, $SO$, $QP$ are all coincident.

Perhaps someone else will see how to prove that first parallelism geometrically.


Meanwhile, here's a vectorial proof of that parallelism and of the theorem as a whole, for anyone who wants an algebraic approach.

Define $b,c,d,e,f$ as vectors, and $t,u$ as scalars such that

  • $b$ goes from $O$ to $S$
  • $c$ goes from $P'$ to $Q'$, $d$ goes from $R'$ to $Q'$, $tc$ goes from $O$ to $R'$.
  • $e$ goes from $R$ to $Q$, $\ f$ goes from $P$ to $Q$, $\ ue$ goes from $O$ to $P$.

Now we record several scalar facts: \begin{align} PQR \text{ are vertices of a square}&: \ \ e.f = 0,\ e.e = f.f & (1)\\ P'Q'R' \text{ are vertices of a square}&: \ \ c.d = 0,\ c.c = d.d & (2)\\ c+f \text{ rotated }90^o \text{ gives }d+e&: \ \ \ \ \ \ \ \ \ \ \ \ |c+f|=|d+e| &(3)\\ P' \text{ lies on }PQ&: (tc+d-c).e=ue.e &(4)\\ R \text{ lies on }Q'R'&: (ue+f-e).c=tc.c &(5)\\ S \text{ lies on }P'Q'&: \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b.d=d.d &(6)\\ S \text{ lies on }QR&: \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b.f=f.f &(7)\\ 1+2+3 &: \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c.f=d.e & (8)\\ 4+5+8 &: \ \ \ \ \ \ \ \dfrac{t}{c.e+e.e}=\dfrac{u}{c.c+c.e} & (9) \end{align} This is enough to prove that $R'P=ue-tc$ and $Q'Q=ue+f-tc-d$ are both perpendicular to $c+e$, and therefore parallel as for the above use of Desargues's theorem. Meanwhile, proceeding vectorically, we check that $$(d.e)b=(e.e)c + (c.c)e$$ because both sides have the same dot product with $d$ (by 2,6), and both sides have the same dot product with $f$ (by 1,7,8), and we assume that $d$ and $f$ span the plane. We conclude that $$\frac{c.c}{b.c}(c.e+e.e)=d.e=\frac{e.e}{b.e}(c.c+c.e)\ \ (10)$$ So multiplying (9) and (10) gives $$\frac{c.c}{b.c}\ t=\frac{e.e}{b.e}\ u =: x$$ Finally, let the point $X$ be $xb$ away from $O$. Then:

  • $xb.c=tc.c$, so $X$ is on $Q'R'$;
  • $xb.e=ue.e$, so $X$ is on $PQ$;
  • by construction $X$ is on $OS$;
  • so $Q'R'$, $PQ$, and $OS$ are coincident, as desired.

Although interesting, this theorem is not new. I recently found the article "Squares Inscribed in Angles and Triangles" on JSTOR.

enter image description here

Bailey, Herbert, and Duane Detemple. “Squares Inscribed in Angles and Triangles.” Mathematics Magazine, vol. 71, no. 4, 1998, pp. 278–284. JSTOR, JSTOR, www.jstor.org/stable/2690699.

It would be nice if I am the person who discovered the fact, but since the theorem was published in 1998, the credit does not belong to me.

  • $\begingroup$ That someone discovered the fact before you does not diminish your accomplishment. The late, great Steve Fisk called this kind of thing a mere "accident of time". :) $\endgroup$
    – Blue
    Oct 19, 2018 at 0:01
  • 1
    $\begingroup$ Thank you for your encouragement. $\endgroup$
    – Larry
    Oct 19, 2018 at 0:06
  • 1
    $\begingroup$ The reference you found shows that the collinearity property holds for arbitrary squares inscribed in the two sides of an angle; the third side of the triangle is a distraction (especially since there's no concurrence). I'm a little surprised I didn't see this earlier: My proof never used the third components in my ratios $\cot C:\pm 1:\cot A$ and $\cot C:\pm 1:\cot B$. I'll refine that proof. In fact, I can generalize so that the result applies not merely to inscribed squares, but inscribed similar parallelograms. (My ostensibly-"new" result may generalize in the same way.) $\endgroup$
    – Blue
    Oct 19, 2018 at 2:15

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