In the triangle $ABC$ the side $AC = \tfrac {1} {2} (AB + BC) $, $BL$ is the bisector $\angle ABC$, $K, \, \, M $ - the midpoints of the sides $AB$ and $BC$, respectively. Find the value $\angle KLM$ if $\angle ABC = \beta$
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Tags: geometry, angles, angle bisector
superagh
07.08.2020 05:41
parmenides51 wrote: In the triangle $ABC$ the side $AC = \tfrac {1} {2} (AB + BC) $, $BL$ is the bisector $\angle ABC$, $K, \, \ , M $ - the midpoints of the sides $AB$ and $BC$, respectively. Find the value $\angle KLM$ if $\angle ABC = \beta$ Is there a typo with the commas?
superagh
07.08.2020 05:51
My little progress: Label $CM=MB=x$ and $AK=KB=y$. Then, $AC=x+y$, and by angle bisector theorem, $AL=y$ and $CL=x$.
rafaello
07.08.2020 12:16
By the angle bisector theorem $\frac{AK}{CM}=\frac{AB}{BC}=\frac{AL}{CL}$ and since $AL+LC=AC = \tfrac {1} {2} (AB + BC) =AK+CM,$ thus $\frac{AC-CM}{CM}=\frac{AC-CL}{CL}\implies LC=CM$, thus also $AL=AK$.
By isosceles triangles, $\angle ALK=\frac{180^{\circ}-\angle BAC}{2}$ and $\angle MLC=\frac{180^{\circ}-\angle ACB}{2}$.
We conclude that $$\angle KLM=180^{\circ}-\angle ALK-\angle MLC=180^{\circ}-\frac{180^{\circ}-\angle BAC}{2}-\frac{180^{\circ}-\angle ACB}{2}=\frac{\angle BAC+\angle ACB}{2}=\frac{180^{\circ}-\angle ABC}{2}=90^{\circ}-\tfrac {1} {2} \beta.$$