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$XYZ$ is a triangle in which $\angle X$ is obtuse. A point $P$ is taken inside the triangle and $XP$, $YP$, $ZP$ are produced to meet the sides $YZ$, $ZX$, $XY$ at the points $K$, $L$, $M$, respectively. Suppose that $PL = PM$.

Find the angles of triangle $XYZ$, given that

  • $XK$, $YL$, $ZM$ are the angle bisectors of triangle $XYZ$
  • and that $2\;XK = YL$.
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  • $\begingroup$ I proved that XYZ is an isosceles triagle; XZ=XY. That means the angle bisector of X will be Perpendicular to YZ (and median). $\endgroup$ – Lama Jan 6 '16 at 20:43
  • $\begingroup$ After this I get confused. . I dont know how to make pythagoras law and sine law usuful in sovling my problem! $\endgroup$ – Lama Jan 6 '16 at 20:54
  • $\begingroup$ I placed a drawing in the answer section that meets the above requirements without being isosceles triangle. Could you provide the proof for your claim? $\endgroup$ – Moti Jan 9 '16 at 2:56
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hint: make $LQ \perp YZ , Q$ is on $YZ$, find $\dfrac{LQ}{XK}$

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The attached drawing show a case where the above requirements are met but I am not sure there is a single solution

enter image description here

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