In an acute-angled triangle $ABC, D$ is the foot of the altitude from $A$. Let $D_1$ and $D_2$ be the symmetric points of $D$ wrt $AB$ and $AC$, respectively. Let $E_1$ be the intersection of $BC$ and the line through $D_1$ parallel to $AB$ . Let $E_2$ be the intersection of$ BC$ and the line through $D_2$ parallel to $AC$. Prove that $D_1, D_2, E_1$ and $E_2$ on one circle whose center lies on the circumscribed circle of $\vartriangle ABC$.
Problem
Source: 2020 Dutch BxMO TST p2
Tags: geometry, Concyclic, cyclic quadrilateral
Com10atorics
20.12.2020 01:52
Wlog $AB<AC$. Let $O$ be the circumcenter of $(ABC)$ and $A'$ the antipode of point $A$ in this circle. First see that $AD_1=AD=AD_2$. And now $\angle OAD_1=\angle OAB+\angle BAD_1=\angle OAB+\angle BAD$ $\angle OAD_2=\angle CAD_2+\angle CAO=\angle CAD+\angle CAO$ So $AO$ bisects $\angle D_1AD_2$ and therefore $A'D_1=A'D_2$. From $BD_1=BD$ we get $\angle BD_1E_1=90-\angle BD_1D=\angle D_1BA=\angle B=\angle BE_1D_1$ which implies that $BD_1=BE_1$. So we have $A'B\perp AB\Rightarrow A'B\perp D_1E_1\Rightarrow A'D_1=A'E_1$ and similarly $A'D_2=A'E_2$. So $(D_1D_2E_1E_2)$ is a circle with circumcircle $A'$. P.s. it can be done with complex numbers too.
Basu_Dev
06.11.2024 06:36
Here, is my solution. I fcked up with the writing but It works ig.
Attachments:
Problem_3.pdf (181kb)
Thapakazi
06.11.2024 15:40
Let $E_3$ be the intersection of $D_1E_1$ and $D_2E_2$. It is clear that $E_3D_2DD_1$ cyclic with center $A$. Therefore notice that \[\angle E_3D_1D_2 = \frac{\angle E_3AD_2}{2} = 90 - \angle D_3E_2 = \angle D_2E_2E_1\] which implies $E_1D_1D_2E_2$ cyclic. Now notice that $E_1$ and $E_2$ are the reflections of $D$ across $B$ and $C$ respectively. Consider the perpendicular bisectors of $D_1E_1$ and $D_2E_2$ and let it intersect at $A'$. As $BE_1 = BD_1$ and $CD_2 = CE_2$ both of these bisectors pass through $B$ and $C$. Furthermore, this implies that $\angle ABA' = \angle ACA' = 90$ which gives us that $ABA'C$ cyclic as needed.