The line $\ell$ parallel to the side $BC$ of the triangle $ABC$, intersects its sides $AB,AC$ at the points $D,E$, respectively. The circumscribed circle of triangle $ABC$ intersects line $\ell$ at points $F$ and $G$, such that points $F,D,E,G$ lie on line $\ell$ in this order. The circumscribed circles of the triangles $FEB$ and $DGC$ intersect at points $P$ and $Q$. Prove that points $A, P$ and $Q$ are collinear.
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Tags: geometry, circumcircle, parallel, Kharkiv, collinear
TaDucDuy
22.03.2020 10:58
Solution:Let S be the intersection of BF and CG then S lies on the radical axis of circles (BEF),(CGD).Let J be the intersection of AS and DE,we will prove that J also lies on the radical axis of (BEF),(CGD).Using Menelaus's theorem for $\triangle EGC$ with $\overline{A,J,S}$,we get $\frac{\overline{JE}}{\overline{JG}}\cdot\frac{\overline{SG}}{\overline{SC}}\cdot\frac{\overline{AC}}{\overline{AE}}=1\Longrightarrow\frac{\overline{JG}}{\overline{JE}}=\frac{\overline{SG}}{\overline{SC}}\cdot\frac{\overline{AC}}{\overline{AE}}=\frac{\overline{SF}}{\overline{SB}}\cdot\frac{\overline{AB}}{\overline{AD}}$ (notice $DE||BC||FG$) (1).Similarly,using Menelaus's theorem for $\triangle FBD$ with $\overline{A,J,S}$,we get $\frac{\overline{SF}}{\overline{SB}}\cdot\frac{\overline{AB}}{\overline{AD}}=\frac{\overline{JF}}{\overline{JD}}$ (2).From (1),(2),it's follow that $\frac{\overline{JG}}{\overline{JE}}=\frac{\overline{JF}}{\overline{JD}}\Longrightarrow\overline{JG}\cdot\overline{JD}=\overline{JF}\cdot\overline{JE}$ ,which means that J lies on the radical axis of (BEF) and (CGD) so JS or the line passing through 3 points A,J,S is the radical axis of (BEF) and (CGD).Therefore 3 points A,J,S are collinear.We done !
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ironball
22.03.2020 19:01
$A,P,Q$ are collinear $\Leftrightarrow$ point $A$ lies on the radical axis of $\odot(BFE)and\odot(CGD) \Leftrightarrow AX\cdot AB=AY\cdot AC$ Let$\frac{AD}{AB}=\lambda$;$AH\perp BC$;$B’$ be the reflection of $B$ through $H$ and $AA’ \parallel BC$;$A’C\cap FG=J$;$AB’\cap FG=I$ $FD\cdot DE=BD\cdot DX\Rightarrow FD\cdot \lambda a=(1-\lambda)c\cdot DX\Rightarrow DX=\frac{\lambda a\cdot FD}{(1-\lambda)c}\Rightarrow AX=\lambda c-\frac{\lambda a\cdot FD}{(1-\lambda)c}$ The same,$AY=\lambda b-\frac{\lambda a\cdot EG}{(1-\lambda)b}$ Therefore,to prove $AX\cdot AB=AY\cdot AC$ is to prove $c[\lambda c-\frac{\lambda a\cdot FD}{(1-\lambda)c}]=b[\lambda b-\frac{\lambda a\cdot EG}{(1-\lambda)b}]$ $\Leftrightarrow ……$ $\Leftrightarrow (1-\lambda)(b^2-c^2)=a(EG-FD)$ $\Leftrightarrow (1-\lambda)(a^2+c^2-2ac\cdot cosB-c^2)=a(EG-FD)$ $\Leftrightarrow (1-\lambda)a(a-2c\cdot cosB)=a(EG-FD)$ $\Leftrightarrow (1-\lambda)(a-2c\cdot cosB)=EG-FD$ $\Leftrightarrow (1-\lambda)B’C=EG-DF$ $\Leftrightarrow B’C-IE=EG-DF$ Obviously,$B’C-IE=EG-DF=EJ$ So we are done.
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