Let $ABC$ be a triangle with $AB < AC$ inscribed in $(O)$. Tangent line at $A$ of $(O)$ cuts $BC$ at $D$. Take $H$ as the projection of $A$ on $OD$ and $E,F$ as projections of $H$ on $AB,AC$.Suppose that $EF$ cuts $(O)$ at $R,S$. Prove that $(HRS)$ is tangent to $OD$
parmenides51 wrote:
Let $ABC$ be a triangle with $AB < AC$ inscribed in $(O)$. Tangent line at $A$ of $(O)$ cuts $BC$ at $D$. Take $H$ as the projection of $A$ on $OD$ and $E,F$ as projections of $H$ on $AB,AC$.Suppose that $EF$ cuts $(O)$ at $R,S$. Prove that $(HRS)$ is tangent to $OD$
Saniva_Rakib_Soha wrote:
Let $ABC$ be an acute-angled triangle with $AB>AC$, circumcircle $S$ and $M$ be the midpoint of $BC$. A point $N$ lies inside of $ABC$ so that $DE$ is perpendicular to $AM$, where $D$ and $E$ are the feet of the altitudes from $N$ to $AB$ and $AC$,respectively. The circumcircle of $ADE$ meets $S$ again at $L$ and $K$ is the intersection point of $AL$ and $DE$. Line $AN$ meets $S$ again at $F$. If $N$ coincides with the midpoint of $AF$, show that $KA=KF$.
In Problem $1$, let $(AEF)\cap (ABC)=P$ and $AP\cap EF=T$. Since $N$ in Problem $2$ is $T$ in Problem $1$, we get in Problem $1$ $TA=TG$ where $G$ is intersection of another tangent to $(ABC)$ from $D$ with $(ABC)$. So $D-T-H$ are collinear $\implies TE\cdot TF=TH^2=TP\cdot TA=TR\cdot TS \implies TR\cdot TS=TH^2$. So $(RHS)$ tangent to $TH$ at $H$. So we are done!