Let $ABC$ be an isosceles triangle. Suppose that points $K$ and $L$ are chosen on lateral sides $AB$ and $AC$ respectively so that $AK = CL$ and $\angle ALK + \angle LKB = 60^o$. Prove that $KL = BC$.
Problem
Source:
Tags: geometry, isosceles, equal segments, angles
Tsikaloudakis
19.10.2021 14:41
$\begin{gathered} \left. \begin{gathered} LD//AK \hfill \\ LE// = BC \hfill \\ \end{gathered} \right\}\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} \left\{ \begin{gathered} KD = KB \hfill \\ DL = BE \hfill \\ K\hat DL = K\hat BE \hfill \\ \end{gathered} \right.\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} tr.DKL = tr.BEK = tr.KAL\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} E\hat KL = 60^O \mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} (1) \hfill \\ BELD,isoskeles\mathop {}\limits_{}^{} \mathop {}\limits_{}^{} trapezioum \Rightarrow \mathop {}\limits_{}^{\mathop {}\limits_{}^{} } \hat L_1 = \hat E_1 \hfill \\ \left. \begin{gathered} \hat L_1 = \hat E_1 \hfill \\ B\hat EK = D\hat LK \hfill \\ \end{gathered} \right\}\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} \hat L_2 = \hat E_2 = 60^O \mathop {}\limits_{}^{} \mathop \Rightarrow \limits^{(1)} \mathop {}\limits_{}^{} KL = EL = BC \hfill \\ \end{gathered} $
Attachments:
