# What's wrong with this problem?

A regular hexagon $ABCDEF$ is given and also two points $K$ and $L$ on its sides $AB,BC$ respectively, such that $\angle KEL=15^{\circ}$. Show that EL bisects the angle $\angle KLC$.

Having made the shape using geogebra, it turns out there's something wrong. Could someone please find what's wrong and suggest what the correct problem is?

Also, if you do find the correct one, do not write a solution

• Just tried it on sketchpad and I agree. EL nowhere near bisects angle KLC, no matter where K is placed on side AB (in such a way that L ends up on side BC keeping angle KCL at 15). It would be interesting if anything constant held here, as the resulting angles vary depending on where K is on AB. – coffeemath Dec 1 '13 at 21:29

## 2 Answers

Since @Matheo has lifted the moratorium on solutions, here's mine.

Reflect $A$ and $C$ across $\overline{EK}$ and $\overline{EL}$, respectively, to get $A^\prime$ and $C^\prime$, both of which are on the circle with center $E$ through $A$ and $C$. Since $|\angle AEC| = 60^\circ$ and $|\angle KEL| = 30^\circ$, we have that $$|\angle A^\prime EK| + |\angle C^\prime EL| = |\angle AEK| + |\angle CEL| = 60^\circ - 30^\circ = |\angle KEL|$$ so that $A^\prime$ and $C^\prime$ must coincide.

By reflection, $\overline{EK}$ and $\overline{EL}$ are respective bisectors of $\angle AKA^\prime$ and $\angle BLB^\prime$. To see that this in fact solves the problem, observe that $\overline{KL}$ contains $A^\prime=C^\prime$: angles $\angle EAB$ and $\angle ECB$ are right, whereupon their reflections, $\angle EA^\prime K$ and $\angle EC^\prime L$, are also right, so that points $K$, $L$, and $A^\prime = C^\prime$ are collinear. $\square$

Note that there's nothing special about the hexagon here. The same phenomenon appears whenever $A$, $B$, $C$ are three consecutive vertices of any regular $2n$-gon, with $E$ the vertex opposite $B$, provided $|\angle KEL| = 90^\circ/n$.

AHA. By messing around on sketchpad, I found if one changes the required angle $KEL$ to be 30 degrees instead of 15, the result holds, at least using the sketchpad drawing. And the value of the two angles in which $EL$ bisects angle $KLC$ vary as the point $K$ is moved on side $AB$, keeping $L$ on side $BC$ so as to have the required angle 30 degrees.

This seems like an interesting result, and I'll likely try for a proof myself. Of course I won't put it here as per your request, which I agree with -- don't spoil the fun...

• @Blue -- Too bad OP doesn't want to see that proof, as I'd like to see it. OTOH I'd still like to go for a proof myself... – coffeemath Dec 2 '13 at 5:41
• If I do find the proof, I will let you know so you can share your solutions with us. :) – Matheo Dec 2 '13 at 12:31
• Okay, job done. It's quite easy actually. – Matheo Dec 2 '13 at 17:34
• @Blue how did you prove it, if I may ask? – Matheo Dec 2 '13 at 21:27