Rectified, Halved, Sheared, Eyeballs Still Surprise

What is this about?


Similar triangles $ACG$ and $DBH$ share the base line $AB.$ $I=CG\cap AH,$ $J=BG\cap DH,$ $K=DG\cap AH,$ $L=BG\cap CH.$

Eyeballs with similar triangles - problem

Prove that $IJ\parallel AB\parallel KL.$


The fact that $KL\parallel AB$ has a projective analogue that is easily solved with Thales' theorem.

Concerning $IJ$ also being parallel to $AB,$ note that if the triangles $ACG$ and $DBH$ are equal, the claim is rather simple.

Eyeballs with similar triangles - solution for IJ||AB

Triangles $ACI$ and $HGI$ are similar, implying $\displaystyle\frac{CI}{GI}=\frac{AC}{GH}.$ Triangles $BDJ$ and $GHJ$ are also similar, implying $\displaystyle\frac{DJ}{HJ}=\frac{BD}{GH}.$ But if $\Delta ACG=\Delta DBH,$ $AC=BD$ so that $\displaystyle\frac{CI}{GI}=\frac{DJ}{HJ}.$ By Thales' theorem, $IJ\parallel AB.$

In the general case, choose $D'$ on $AB$ such that $D'B=AC$ and make $D'H'\parallel DH,$ with $H'$ on $BH.$ Define $I'$ and $J'$ accordingly. We just saw that $I'J'\parallel AB.$

Eyeballs with similar triangles - solution for IJ||AB, part 2

From a several pairs of similar triangles,




so that




and (from $\Delta ADH$) making $IJ\parallel AB.$


The is the extension of the Eyeball Theorem Rectified problem first suggested by Machó Bónis $(IJ)$ and then augmented by Dao Thanh Oai $(KL).$

Leo Giugiuc came up with a short analytic solution.

Hubert Shutrick has been insisting that the whole statement is of projective nature. He came up with a convincing dynamic illustration that I placed in a separate file.

Related material

Problems with Ophthalmological Connotations

  • The Eyeball Theorem
  • Eye-to-Eye Theorem I
  • Eye-to-Eye Theorem II
  • The Squinting Eyes Theorem
  • Eyeballing a ball
  • Praying Eyes Theorem
  • Focus on the Eyeball Theorem
  • Eyeball Theorem Rectified
  • Bespectacled Eyeballs Extension
  • Shedding Light on the Ball for Eyeballing
  • Eyeballs Projected
  • Archimedean Siblings out of Wedlock, i.e., Arbelos
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