In an acute-angled triangle $ABC$ let $BL$ be the bisector, and let $BK$ be the altitude. Let the lines $BL$ and $BK$ meet the circumcircle of $ABC$ again at $W$ and $T$, respectively. Given that $BC = BW$, prove that $TL \perp BC$.
Problem
Source: Caucasus MO 2024, Seniors P2
Tags: geometry
15.03.2024 17:16
15.03.2024 18:57
Let $TL$ meet $BC$ at $P$. Firstly, let the angles of the triangle be $\alpha, \beta, \gamma$ for $\angle A,\angle B$ and $\angle C$ respectively. $BL$ is angle bisector $\implies \angle ABW= \beta/2$. Quadrilateral $AWCB$ is cyclic $\implies \angle ABW=\angle ACW= \beta/2$. $\angle BAC= \angle BWC= \alpha$. Given, $BW=BC \implies \angle BCW= \angle BWC= \alpha$. $\implies \angle BCA+\angle ACW= \alpha$. But $\angle BCA=\gamma \implies \angle ACW= \alpha-\gamma$. Now, equating the two values of $\angle ACW$, we get that $\beta/2=\alpha-\gamma \implies \alpha=\beta/2+\gamma$. $\angle CLW= \angle WBC+\angle BCL=\beta/2+\gamma=\alpha$. $\angle BAC=\alpha$. Therefore, we get that $|delta ABL$ is isosceles, with $AB=BL$. As $BK$ is altitude of an isosceles triangle, $AK=KL$. Also, $\angle TKL=\angle TKA=90^{\circ}$ $\implies$ by $SAS$, $\Delta TAK \cong \Delta TLK$ $\implies \angle ATK= \angle LTK$ As $ATCB$ is cyclic, we get that $\angle ATB=\angle ATK=\angle ACB= \gamma$ $\implies \angle ATK=\angle LTK=\gamma$. In $\Delta CKB$, we have $\angle CKB= 90^{\circ}$, and $\angle KCB=\gamma \implies \angle KBC= 90^{\circ}-\gamma$. Therefore, in $\Delta TBP$, by angle sum, we get $\boxed{\angle TPB=90^{\circ}}$, and we're done!
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24.03.2024 21:44
Toss the figure on the complex plane. Without loss of generality we may assume $A,B,C$ are on the unit circle. Further we may assume $B = 1, C = c$. This implies that $W = \overline{c} = \frac 1c$. This implies that $A = \frac{1}{c^3}, T = -\frac{1}{c^2}$. Calculating $L = AW \cap BC$ we get that $L = c + 1 - \frac{1}{c^2}$. Thus, $\frac{(L - T)}{(C-B)} = \frac{c+1}{c-1}$ which is purely imaginary, and we are done.