0
$\begingroup$

$XYZ$ 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 $2\;XK = YL$.
$\endgroup$
3
  • $\begingroup$ 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
0
$\begingroup$

hint: make $LQ \perp YZ , Q$ is on $YZ$, find $\dfrac{LQ}{XK}$

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.