Blastzit wrote: Let $\Delta ABC$ be an acute triangle with incenter $I$ and orthocenter $H$. $AI$ meets the circumcircle of $\Delta ABC$ again at $M$. Suppose the length $IM$ is exactly the circumradius of $\Delta ABC$. Show that $AH\geq AI$. Let $O$ be the circumcenter of $\odot{ABC}$.By Fact 5 $MI=MB=MC=MO=OB=OC$.Hence $OMC$ and $OMB$ are equilateral triangles.Hence $\angle{BAC}=2\cdot \angle{MAC}=60^{\circ}$.Also $AI\cdot R=AI\cdot IM=R^2-OI^2=2Rr \implies AI=2r$.Now \[R=2R\cos{60^{\circ}}=AH\geq AI=2r\]which is just Euler's Theorm so we are done.

$M$ -center of $\odot IBC$ which in this case is a reflection of $\odot ABC$ about $BC$, so it goes through $H$. $$AH+HM\geq AM \implies AH \geq AI$$

Identity:

. As $MO=MI=MB=MC$ we get that $,O,I,B,C$ lie on a circle $\omega$ . So $\angle BOC=\angle BIC$, using above identity we have $ \angle BHC=\angle BIC$ and hence $H\in\omega$. Now $AH+HM\geq AM \implies AH\geq AI$.

Let $I_A,I_B,I_C$ be the $A,B,C$ excenters of $\Delta ABC$.We have $I_A$ is the reflection of $I$ in $M$.So $II_A=2R$.Again if $O_1$ is the centre of $\Delta I_AI_BI_C$ then $I_AO_1=2R$ as circumcircle of $\Delta ABC$ is nine point circle of $\Delta I_AI_BI_C$.So $II_A=I_AO_1$.As $I$ is the orthocentre of $\Delta I_AI_BI_C$ and $O_1$ is the circumcenter of $\Delta I_AI_BI_C$ and the are of equal length so $\angle{I_CI_AI_B}=60^{\circ}$ .So $\angle{I_CI_AI_B}=90^{\circ}-\angle{BAC}/2=60^{\circ}$ so $\angle{BAC}=60^{\circ}$.So $\angle{BHC}=120^{\circ}$. Now note that $\angle{BHC}+\angle{BI_AC}=120^{\circ}+60^{\circ}=180^{\circ}$ so $BHICI_A$ is concyclic.By fact 5 $M$ is the centre of this circle.So $MH=MI$.Now $AH+HM\geq AM \implies AH\geq AI$.