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Suppose $A-D-B$ and $A-E-C$ are such that segment $DE$ is perpendicular to line $AE$ at $E$ and segment $BC$ is perpendicular to line $AC$ at $C$. Let $x_1$ be the length $DE$ and $x_2$ is the length $BC$. Prove $x_1<x_2$. Note this is in absolute geometry(neutral geometry) so no access to parallel postulate or midpoint connector theorem (also no similar triangles).enter image description here

My idea so far is that I know by the exterior angle inequality applied to the triangle $ADE$ we have that $\angle E$ is greater then $\angle A$. From that I want to then say that it will follow that $x_2>x_1$ from another inequality. I would like to use scalene inequality but not sure how that would work with this construction. Any help would be appreciated.

Edit: Another idea I have is that triangles have non zero defects. So if you connect $D$ to $C$ we get $3$ triangles in the figure. Thus $\delta( \triangle BAC)= \delta(\triangle DAE)+ \delta(\triangle DEC) +\delta(\triangle DCB)$ Then since all $\delta(\triangle) >0$ we get that $180- \sigma(\triangle BAC)> 180- \sigma(\triangle DAE)$ Thus can conclude that $\angle ABC< \angle ADE$ Now from here not sure what I can do my idea was hinge theorem but doesn't work with the sides. Also not quite the right angles I want anyways.

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    $\begingroup$ The triangles are similar (both have one $90^°$deg angle and one Shared angle) so $AC/AE=BC/DE$, we can rewrite the length of $AC$ with some positive constant $c$: $AC=AE+c$ so we get $BC/DE=(AE+c)/AE=1+c/AE>1\implies BC/DE>1\implies BC>DE$(when I write XY, I refer to the length of it) $\endgroup$ – Holo Apr 9 '18 at 4:23
  • $\begingroup$ In absolute geometry you can't use similar triangles though $\endgroup$ – HighSchool15 Apr 9 '18 at 20:38
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🔺 ADE & ABC are similar because angle DEA & BCA are 90 deg. and angle DAE & angle BAC are equal so we can say x1/x2 = AE/AC and AE

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  • $\begingroup$ In absolute you can't use similar triangles $\endgroup$ – HighSchool15 Apr 9 '18 at 20:38

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