Given an acute triangle $ABC$ satisfying $\angle ACB<\angle ABC<\angle ACB+\dfrac{\angle BAC}{2}$. Let $D$ be a point on $BC$ such that $\angle ADC=\angle ACB+\dfrac{\angle BAC}{2}$. Tangent of circumcircle of $ABC$ at $A$ hits $BC$ at $E$. Bisector of $\angle AEB$ intersects $AD$ and $(ADE)$ at $G$ and $F$ respectively, $DF$ hits $AE$ at $H.$ a) Prove that circle with diameter $AE,DF,GH$ go through one common point. b) On the exterior bisector of $\angle BAC $ and ray $AC$ given point $K$ and $M$ respectively satisfying $KB=KD=KM$, On the exterior bisector of $\angle BAC$ and ray $AB$ given point $L$ and $N$ respectively satisfying $LC=LD=LN.$ Circle throughs $M,N$ and midpoint $I$ of $BC$ hits $BC$ at $P$ ($P\neq I$). Prove that $BM,CN,AP$ concurrent.
Problem
Source: Vietnam TST 2016
Tags: geometry, circumcircle
26.03.2016 12:32
Any solutions?
07.04.2017 18:59
Any solution
30.04.2019 06:23
Here is my solution for this problem Solution a) Let $S$ be center of ($ADE$); $W$ $\equiv$ $AF$ $\cap$ $BC$; $J$ be Miquel point of completed quadrilateral $AFDE.HW$ We have: ($WC$; $WA$) $\equiv$ ($DC$; $DA$) + ($AD$; $AF$) $\equiv$ ($CA$; $CB$) + $\dfrac{(AB; AC)}{2}$ + ($EC$; $EF$) $\equiv$ ($CA$; $CB$) + $\dfrac{(AB; AC)}{2}$ + $\dfrac{(EC; EA)}{2}$ $\equiv$ ($CA$; $CB$) + $\dfrac{(AB; AC)}{2}$ + $\dfrac{(BC; BA) - (CA; CB)}{2}$ $\equiv$ $\dfrac{(BC; BA) + (CA; CB) + (AB; AC)}{2}$ $\equiv$ $\dfrac{\pi}{2}$ (mod $\pi$) So: $AF$ $\perp$ $BC$ at $W$ Since: $A$, $F$, $D$, $E$ lie on a circle then by Brocard theorem, we have: $G$ is orthocenter of $\triangle$ $SHW$ Hence: $SG$ $\perp$ $HW$ at $J$ So: $J$ $\in$ $\bigodot(HG)$ But: $J$ be Miquel point of completed quadrilateral $AFDE.HW$ then: $J$ $\in$ $\bigodot(FD)$, $J$ $\in$ $\bigodot(EA)$ Hence: $\bigodot(FD)$, $\bigodot(EA)$, $\bigodot(HG)$ have a common point $J$ b) Let $U$ be intersection of $A$ - internal bisector of $\triangle$ $ABC$ with $BC$; $Q$ $\equiv$ $MN$ $\cap$ $BC$ Since: $AK$ is external bisector of $\widehat{BAC}$ and $BK$ = $KM$ then: $A$, $K$, $B$, $M$ lie on a circle Similarly: $A$, $L$, $C$, $N$ lie on a circle We have: ($DC$; $DM$) $\equiv$ $\dfrac{(KB; KM)}{2}$ $\equiv$ $\dfrac{(AB; AC)}{2}$ $\equiv$ ($AU$; $AC$) (mod $\pi$) So: $M$ $\in$ ($ADU$) Similarly: $N$ $\in$ ($ADU$) Then: ($ND$; $NA$) $\equiv$ ($UD$; $UA$) $\equiv$ ($CB$; $CA$) + $\dfrac{(AC; AB)}{2}$ $\equiv$ ($DA$; $DC$) $\equiv$ ($DA$; $DB$) (mod $\pi$) or $AD^2$ = $\overline{AN}$ . $\overline{AB}$ Similarly: $AD^2$ = $\overline{AM}$ . $\overline{AC}$ So: $\overline{AN}$ . $\overline{AB}$ = $\overline{AM}$ . $\overline{AC}$ = $AD^2$ or $B$, $C$, $M$, $N$ lie on a circle Then: $\overline{QB}$ . $\overline{QC}$ = $\overline{QN}$ . $\overline{QM}$ = $\overline{QP}$ . $\overline{QI}$ or ($BCPQ$) = $-$ 1 Hence: $AP$, $BM$, $CN$ concurrent
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