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Let $A’$ be the antipode of $A$ w.r.t $(ABC)$. Note that $K,H,M,A’$ are collinear. Note that it is enough to prove that the antipode of $A$ w.r.t $(ADE)$ lies on $HM$. Let $\Delta PQR$ be the orthocentre triangle of $\Delta ABC$ and $X = HM \cap A-$anglebisector. Note that $\frac{HX}{XA’} = \frac{AH}{AA’} = \frac{HQ}{HC} = \frac{QE}{EC}$ by repeated use of angle bisector theorem. Also, $HQ \parallel CA’$. Thus $XE \perp AC$. Similarly (or by Simson’s line), we have that $XD \perp AB$. Thus, the claim holds true. $\blacksquare$.

Another way to show this is to use the fact that $H$ is the $K$-Humpty point of $\triangle KBC$. Then, by the spiral similarity lemma, we just need to show that $$\frac{BD}{CE}=\frac{KB}{KC}=\frac{HB}{HC} \Leftrightarrow \frac{BD}{BH}=\frac{CE}{CH} \Leftrightarrow \angle BHD=\angle CHE$$where the last statement follows by using sine law in $\triangle BHD$ and $\triangle CHE$, and because $\angle BDH=\angle CEH$. Thus, it suffices to show that $DE$ is the external angle bisector of $\angle BHC$; or equivalently that the internal angle bisector of $\angle BHC$ and $\angle BAC$ are parallel to each other. But this is a well known property of an orthocentric system. Hence, done. $\blacksquare$

Let, $MH\cap AO=R$ which lies on $(ABC)$ It's enough to prove that $KA$ is the external angle bisector of $\angle{DKE}$ By Reim's theorem, $DE||BP||QC$ where $P=KE\cap(ABC)$ and $Q=KD\cap(ABC)$ Easy angle chasing gives that, $CA$ is the internal angle bisector of $\angle{QCP}$ which implies that arc$AQ=$arc$PA$ implying arc$RQ=$arc$RP$ So we are done...

This literally took me like 2.5 hours, so here we go. Also note that i think i set $D$ as $E$ and $E$ as $F$, so this might be kinda confusing. First, let $Q$ be the foot of the perpendicular from $B$ to $AC$ and let $R$ be the equivalent for $C$. It’s well-known (or just look at the solutions above) that $APRHQ$ is cyclic. Now, perform a $\sqrt{AR \cdot AB}$ inversion around $A$. Note that $R$ gets sent to $B$, $Q$ gets sent to $C$, and vice versa. If we let $Y$ be the point on $BC$ such that $Y$ bisects $\angle{BAC}$, we have that the line $EF$ is sent to the circle with diameter $Y$. Thus, if $E’$ and $F’$ denote the images of $E$ and $F$ under the inversion respectively, we have that they are the foots of $Y$ onto $AB$ and $AC$. We also have that $H’$ is sent to the foot of the altitude from $A$ to $BC$. Finally, $P$ gets sent to the intersection of $QR$ and $BC$. Thus, it suffices to prove that the image of $P$ lies on the line $E’F’$. Now, with our previous definition of $P$ we had that $(P, H ; R, Q) = -1$. Since inversion preserves cross-ratio, if we define $P’$ as the intersection of $E’F’$ and $BC$, it suffices to prove that $(P’, H’ ; B, C) = -1$. Now, let $B’$ be the reflection of $B$ over $AY$. We have that: \begin{align*} \frac{P’B}{P’C} &= \frac{E’B’}{E’C} \\ &= \frac{F’B}{E’C} \end{align*} Now, since $AF’H’Y’E’$ is cyclic, we have that: \begin{align*} \frac{BF’}{BH’} &= \frac{BY’}{BA} \\ &= \frac{CY’}{CA} \\ &= \frac{CE’}{CH’} \\ \frac{P’B}{P’C} &= \frac{F’B}{E’C} \\ &= \frac{H’B}{H’C} \\ (P’, H’ ; B, C) &= -1 \end{align*} and thus we are done.

This is the necessary and sufficient condition to prove ISL 2005 G5...

No it's very different, read the title.

It’s different, but each problem implies directly in the other.

Based on khina's solution. With same inversion, let $X$ to be the intersection of $RQ$ with $BC$ (or the image of $P$). By brocards, since $RC$ and $BQ$ intersect at $H$, we get $AH$ is the polar of $X$ with respect to $(BRQC)$. Since $AH$ intersects $BC$ and $H'$, we find that $(X,H';C,B)=-1$. Let $X'$ be the intersection of $E'F'$ with $BC$. We claim that $F'C$ and $BE'$ intersect on $AH$. Notice that this suffices to prove the problem as we may use brocards again to prove $(X',H';C,B)=-1$. To prove this, we use Ceva's theorem w.r.t triangle ABC. Since $\frac{E'C}{AE'}=\frac{E'Y\cot(C)}{E'Y\cot{A/2}}=\frac{\cot{C}}{\cot{A/2}}$, $\frac{AF'}{F'B}=\frac{F'Y\cot{A/2}}{F'Y\cot(B)}=\frac{\cot{A/2}}{\cot{B}}$, $\frac{H'B}{CH'}=\frac{AH'cot(B)}{AH'cot{C}}=\frac{cot{B}}{cot{C}}$, we get that $\frac{E'C}{AE'}\frac{AF'}{F'B}\frac{H'B}{CH'}=\frac{\cot{C}}{\cot{A/2}}\frac{\cot{A/2}}{\cot{B}}\frac{cot{B}}{cot{C}}=1$. This proves the above claim. Extension: I seek to prove that $X$, the intersection of $HM$ and the A-bisector is on (ADEP). By the same inversion, it suffices to prove that $X$ gets sent to $N$, the midpoint of $E'F'$. $M$ gets sent to $M'$, the intersection of $(ARQ)$ and $AM$ (since M was on $BC$ and line $BC$ gets send to $(ARQ)$). We see that, by phantom points, it suffices to prove that $M'NP'A$ concyclic, as that would imply that, reversing the inversion, $N'$ is both on the angle bisector of A and line $MP$, which is infact point $X$. With the above proof, $P'$ is on $E'F'$, so <$ANP'$=<$ANE'=90$. It suffices to prove that <$AM'P'=90$ or $M'HP'$ collinear as <$AM'H=90$ since $AH$ is diameter of $(ARQ)$. This is true since $<MH'A=90=<MPA$, so $AMPH'$ concyclic. Under inversion, this means that $M'P'H$ is collinear. This completes the proof.

Let $X$,$Y$,$Z$ be the foots of the altitudes from $A$,$B$,$C$,respectively,and redefine $K$ as the intersection of circles $(ABC)$, $(ADE)$ , $K \neq A$ Claim:The circles $(ABC)$,$(AYZ)$,$(ADE)$ are coaxial

Now,if $O$ and $O'$ are the centers of $(ABC)$,$(AYZ)$ and $A’$ is the antipode of $A$ w.r.t $(ABC)$ note that $ OO'|| A'H$ so $ OO'|| MH$ by the fact that $A',M,H$ are collinear.Also $KA \perp KA'$.But,by the claim $KA$ is the radical axis of $(ABC)$,$(AYZ)$ so $KA \perp OO'$,therefore $KA' || OO'$.Therefore $K \in MH$, so the problem is solved. PS:My first post

Problem wrote: Prove that Nibab can draw a circle I hope Nihab has a steady hand :0

rocketscience wrote: Problem wrote: Prove that Nibab can draw a circle I hope Nihab has a steady hand :0 Nibab*, not Nihab

Let $ \overarc{ADE} \cap \overarc{ABC} : K' $ Let the foot of altitute from $ B $ and $ C$ $Y$ and $Z$. Then, does $\bar{AK'}, \bar{YS}, \bar{BC} $ concurrent?