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$ABCD$ is a square and there is a point $E$ such that $\angle EAB = 15^{\circ}$ and $\angle EBA = 15^{\circ}$. Show that $\triangle EDC$ is an equilateral triangle.

enter image description here

Now there is a proof by contradiction to this problem. I was wondering if there is a pure geometric solution too (no trigonometry)?

rae306
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pirsquare
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5 Answers5

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The best way to show that $\triangle CDE$ is equilateral is to construct an equilateral triangle $\triangle CDE'$ and argue that $\angle ABE'=\angle BAE'=15^\circ$. One can easily see this by looking at the isosceles triangles $\triangle ADE'$ and $\triangle BCE'$.

It is then obvious that $E$ coincides with $E'$.

Here's another proof which is not "backward":

enter image description here

Construct $F$ such that $\angle BCF=\angle BCF=15^\circ$. It follows that $\triangle BEF$ is equilateral since $BE=BF$ and $\angle EBF=60^\circ$.

Also $CF\perp EB$ as $\angle FCB+\angle EBC=90^\circ$. That means the line $CF$ is the perpendicular bisector of $BE$. Thus $CE=CD$.

Quang Hoang
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  • You mean, line $CF$ is the perpendicular bisector of line segment $BE$? (I'm not sure how you got that. I get the "perpendicular" part, though.) – Akiva Weinberger Aug 24 '14 at 17:42
  • @columbus8myhw: Note that I already showed $\triangle BEF$ is equilateral, that mean $F$ is on the perpendicular bisector of $BE$ already. And thanks for spotting out the typo. – Quang Hoang Aug 24 '14 at 17:44
  • Ah, I see. So, $CE=CB$ because $C$ is on the perpendicular bisector, and $CB=CD$ because it's a square... and this leads to $CE=CD$, as claimed. (And, by symmetry, $CE=DE$, and we have $CD=CE=DE$ so $\triangle CDE$ is equilateral.) – Akiva Weinberger Aug 24 '14 at 17:47
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Here is a geometric solution provided by Dr.Shailesh Shirali:

Draw a copy of triangle EAB with DA as base. That is, locate point F inside the square such that triangle FDA is congruent to triangle EAB.

Then angle FAD = angle EAB = 15 deg, so angle FAE = 60 deg. Also, FA = EA. Hence triangle FAE is equilateral, and angle AFE = 60 deg, and FA = FD = FE.

Therefore F is equidistant from A, D, E and so is the circumcentre of triangle DAE.

Therefore by the circle theorems, angle AFE = 2 angle ADE.

But angle AFE = 60 deg. Hence angle ADE = 30 deg, and it follows that angle CDE = 60 deg. So triangle CDE is equilateral.

enter image description here

pirsquare
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  • Great solution. It's easy to follow every step. But, as always with such "tricks": how can one come up with such a solution not knowing the simple solution upfront? What was the train of thoughts / strategy which one could follow to actually come up with such a solution themselves? – BarbaraKwarc Aug 17 '16 at 09:11
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Reflect triangle $AED$ in $AE$ to obtain $AED'$. Then $\angle BAD'$ is $60^{\circ}$. Since $AB=AD=AD'$ this implies that $D'$ is straight below $E$ and therefore $\angle AED=\angle AED'=75^{\circ}$. Then $\angle CDE=75^{\circ}-15^{\circ}=60^{\circ}$.

WimC
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Pirsquare

Let $DC=a~$ and $EC=b$. Using cosine and sine theorems in triangles $EDC$ and $EBC$ , respectively we get that,$~a=2bcosx~$and $~\frac{sin(15+x)}{a}=\frac{sin75}{b}~$.These equations yield $\frac{sin(15+x)}{cosx}=2cos15~$. Finally, $tanx=2-tan15=\sqrt3$. So, $x=60~$.

M.R. Yegan
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Not sure it's what you mean by "proof by contradiction", anyway:

Don't assume you know the angles $EAB$ and $EBA$, but assume instead that the triangle $DCE$ is equilateral.

Then triangles $DEA, CEB$ are isoscele, thus $CEB=CBE=DEA=DAE$. But $EDA=ECB=90°-60°=30°$. Thus $DAE=CBE=75°$. Then $EAB=EBA=15°$.

Since there is only one position of $E$ on the mediatrix of $DC$ such that angle $BEA$ takes on a given value, your proposition is proved.

Jean-Claude Arbaut
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  • Proof by contradiction I meant: let angle EDC =x, angle DEC=(180-2x), angle EDA=(90-x), angle DEA=(x+15). Now suppose that x >60. Then 3x>180 therefore x>(180-2x). Therefore angle DCE>angle DEC. Therefore DE>DC. Also, x+15>75, so in triangle DEA we have angle DEA>angle DAE. Therefore AD>DE. Thus we have AD>DE>DC, an absurdity because AD=DC! Likewise for x<60. So the only option left is to conclude x=60 and this means triangle DEC is equilateral. – pirsquare Aug 25 '14 at 10:37
  • @pirsquare So that' yet another proof. They follow the same idea: use angles to find the solution, but whereas you start from your hypothesis to find a contradiction if $x\neq60°$, I strart with an equilateral triangle to prove $BEA=15°$. – Jean-Claude Arbaut Aug 25 '14 at 10:46