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In triangle $ABC$ the angle $\angle C= 30^\circ$. If $D$ is a point on $AC$,and $E$ is a point on $BC$ such that $AD=BE=AB$.how to prove that $OI=DE$, and how to prove that $OI\perp DE$ where $O$ is the circumcenter, and $I$ is the incenter.

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We have a Cut-the-Knot page for this: :D – Sawarnik Jan 8 '15 at 12:40
up vote 10 down vote accepted

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1. Consider a circle of radius $|OI|$ with center at $I$. Let $OG || BC$ and $OJ || AC$. Also $|KC|=|KA|$.

We can write that $|OJ|=2|KF|=2|AK|-2|AF|=|AC|-(|AB|+|AC|-|BC|)=|BC|-|AB| \ \ ^{*)}$.

As $|AB|=|BE| \Rightarrow |OJ|=|CE|$. Similarly we can prove that $|OG|=|CD|$.

$\angle GOJ =\angle ECD, |OJ|=|CE|, $|OG|=|CD|$ \Rightarrow \triangle GOJ =\triangle ECD $.

Radius of circimcircle of $\triangle GOJ$ is equal to radius of circimcircle of $\triangle ECD \Rightarrow |HE|=|HD|=|OI|$.

$\angle ECD = 30^\circ \Rightarrow \angle EHD=60^\circ \Rightarrow \triangle EHD$ - equilateral and |DE|=|OI|.

enter image description here

2. $OL$ is a tangent line to a circle at point $O$. Then $\angle GJO =\angle GOL$.

As $\triangle GOJ =\triangle ECD \Rightarrow$ $\angle CED =\angle GOL$.

But $GO||BC$ so we have that $OL||DE$

As tangent line $OL\perp OI \Rightarrow DE\perp OI$. And it's true for any $\angle C$.

Both pictures for the case when $|AB|<|BC|$ and $|AB|<|AC|$. Other cases - similarly.

All we need - prove that $\triangle GOJ =\triangle ECD$.

$^{*)}$ UPD To clarify $|OJ|=2|KF|=2|AK|-2|AF|=|AC|-(|AB|+|AC|-|BC|)=|BC|-|AB|$

a. $OK \perp AB$ as $O$ is circumcenter. Thats why $OKWF$ rectangle and $|OJ|=2|KF|$. (see comments).

enter image description here

b. As I is the incenter $\Rightarrow |AF|=|AM|,|CF|=|CN|,|BM|=|BN| \Rightarrow $

$\Rightarrow |AB|+|AC|+|BC| = 2|AF|+|CF|+|CN|+|BM|+|BN|$.

$|CF|+|BM|=|CN|+|BN|=|BC| \Rightarrow$

$\Rightarrow |AB|+|AC|+|BC| = 2|AF|+2|BC| \Rightarrow $

$\Rightarrow 2|AF|=|AB|+|AC|-|BC|$.

It is interesting that points C,I,E,A lies on a circle (and it's true for any $\triangle ABC$). I tried to use this fact to a simpler proof, but I did not succeed. Maybe someone will be lucky.

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Gee, this ain't basic geometry, is it? Anyway, from the beginning I've problems understanding: why $\,|OJ|=2|KF|\,$? A far as I can see, $\,K\,$ is the midpoint of $\,AC\,$ and $\,F\,$ the point of $\,AC\,$ where the perpendicular to it passes through $\,I\,$ and through the point $\,W\,$ of intersection of $\,OJ\,$ with $\,IF\,$ , from which follows that $\,OK\perp AB\,$ , as then $\,OWFK\,$ is a rectangle...But why? – DonAntonio Sep 2 '12 at 2:01
$OK \perp AB$ as O is a circumcenter. $IF \perp AB$ point where inscribe circle tangent AB. Let $OJ$ and $IF$ intersect at $W$. Then $OKFW$ - is a rectangle and $|OJ|/2=|WO|=|KF|$. – Mike Sep 2 '12 at 2:20
Oh, of course. I totally forgot $\,O\,$ is the triangle's circumcenter. Thanks, I'll try to read some more later. – DonAntonio Sep 2 '12 at 2:24
It's nice problem for olympiad. I saw that these triangles are equal, and spent almost half an hour to prove it. Not so hard, as it's seems. – Mike Sep 2 '12 at 2:26

There is not much need of such extravagant constructions. Here is how I have done it:

enter image description here

  1. Note that $\triangle BID \cong \triangle BIC$ by SAS criteria, as we have $BD =BC$ and $\angle BID = \angle BIC$. Similarly, $\triangle CIE \cong \triangle CIB$. Thus, $\angle DIE = 360^{^\circ}- 3\angle BIC=90^{^\circ}- \frac{3\angle A}2=45 ^{\circ}$.
  2. We know that $\angle BOC= 2\angle A = 60 ^{\circ}$, so $\triangle BOC$ is equilateral and hence $BC=OB$. Thus $\triangle OBD$ is isosceles as well, and noting that $\angle OBA=90^{\circ}-\angle C$, we have $\angle ODB= \frac{90^{\circ}+\angle C}2$. As $\angle IDB= \frac{\angle C}2$, we conclude that $\angle ODI = \angle ODB - \angle IDB=45^{\circ}$ and similarly $\angle OEI = 45^{\circ}$.

  3. Thus $DO \perp IE$, and $EO \perp DI$, which means that $O$ is the orthocenter of $\triangle IDE$, and thus we have that $IO\perp DE$. Using the fact that $O$ is the orthocenter, we finally also have $ OI= DE\cdot \cot \angle DIE=DE$.

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