Bariz - Problem

Problem

Source: Greek TST,2012,Pr.2

Tags: geometry, perpendicular bisector, angle bisector

gavrilos

23.05.2016 16:39

Given is an acute triangle $ABC$ $\left(AB<AC<BC\right)$,inscribed in circle $c(O,R)$.The perpendicular bisector of the angle bisector $AD$ $\left(D\in BC\right)$ intersects $c$ at $K,L$ ($K$ lies on the small arc $\overarc{AB}$).The circle $c_1(K,KA)$ intersects $c$ at $T$ and the circle $c_2(L,LA)$ intersects $c$ at $S$.Prove that $\angle{BAT}=\angle{CAS}$.

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FabrizioFelen

23.05.2016 17:40

Let $M\equiv AD\cap \odot (ABC)$ and let $T'\equiv LD\cap \odot (ABC)$ and $S'\equiv KD\cap \odot ABC$ $\Longrightarrow$ since $LA=LD$ $\Longrightarrow$ $DM$ $=$ $DT'$, so from $180^{\circ}-\measuredangle BT'L=\measuredangle BAL=\measuredangle DBL$ we get $\measuredangle BT'D=\measuredangle BDT'$ $\Longrightarrow$ $BD$ $=$ $BT'$ hence $T'$ $=$ $T$ similarly $S'=S$ $\Longrightarrow$ so from $MT$ $=$ $MS$ and $MB=MC$ we get $TS\parallel BC$ $\Longrightarrow$ $\measuredangle BAT$ $=$ $\measuredangle BCT$ $=$ $\measuredangle CBS$ $=$ $\measuredangle CAS$ hence $\measuredangle{BAT}$ $=$ $\measuredangle{CAS}$.

Reynan

23.05.2016 18:55

first $\measuredangle LKD= \measuredangle LKA=\measuredangle LSA=\measuredangle LAS =\measuredangle LKS$ so $K,D,S$ collinear, analog $L,D,T$ collinear $\measuredangle ATL=\measuredangle ASL=\measuredangle LSA = \measuredangle LTS$ so $TL$ is the angular bisector of $STA$ analog $SK$ is the angular bisector of $AST$ so $D$ is the incentre of $\Delta ATS$ so $\measuredangle TAD=\measuredangle SAD\implies \measuredangle{BAT}=\measuredangle{CAS}$

AlastorMoody

19.06.2019 23:43

Let $AD \cap \odot (ABC)=M$ $\implies$ $D$ is orthocenter WRT $\Delta KLM$ $\implies $ $T - D - L$ and $K - D -S$ and since, $MT=MS$ $\implies$ $TS||BC$

EulersTurban

12.03.2020 16:19

So let's start off by extending $AD$,till it hits the circumcircle.Let that point be $M$.Now let $\angle BAT = x$ and $\angle CAS = y$. We know that the quads $BCAK,CBAL$ are cyclic.Thus we know that $\angle BKA = 180- \angle C$ and $\angle CLA = 180 - \angle B$. I will focus on one side. As we know $\angle BKT = \angle BAT = x$,thus we know that $\angle TKA = 180-\angle C - x$.Thus from the diagram we know that $\angle TEA = \frac{1}{2} \angle TKA = 90 - \frac{\angle C + x}{2} $,and from the cyclic quad $TDAE$ we get that $\angle TDA = 180 - \angle TEA = 90 + \frac{\angle C + x}{2} $. If we do the same thing from the other side(that is focus on the cyclic quad $SDAF$ we get that $\angle SDA = 90 + \frac{\angle B + y}{2} $ Now we know that: $$ \angle TDS = 360- \angle TDA - \angle SDA = 360 - 90 - \frac{\angle C + x}{2} - 90 - \frac{\angle B + y}{2} = 180 - \frac{\angle C + \angle B + x +y}{2} = 180 - \frac{180-\angle A + x +y}{2} = 90 + \frac{\angle A - x -y}{2} $$But notice that $\angle TAS = \angle A - x - y $ Thus we can conclude that $D$ is the incenter of the $\triangle TAS$ ( Because of the angle and because $D \in AM$) Thus we can conclude $MT = MS$ Thus we can use this useful lemma:

Now back to the problem. Using this lemma we see that $SC = BT$,and now finally we have proven that $\angle BAT = \angle CAS$......

lazizbek42

27.11.2021 07:32

T,D,L collinear K,D,S collinear

WALKERSkeren

17.11.2022 18:05

lazizbek42 wrote: T,D,L collinear K,D,S collinear The conclusion is easy to be seen in the figure,but how to prove it？