Two of edges' length is given. E is excenter point.

This is geogebra output of a question from my textbook.It can be not to scale by the way.

$B,D,F$ is linear. $E$ is the excenter. $EF$ is perpendicular to $[BF]$. In triangle $\triangle{ABD}$, $BE$ is the bisector of $\angle{ABD}$ and $DE$ is the bisector of $\angle{ADF}$. $|BD|=10$, $|AD|=8$ and $|DF|=5$. Find $|AB|.$

I tried to use bisector properties but i can't get the answer. I think we should use excenter. Any hint will be appreciated.

  • 2
    $\begingroup$ Could you give the exact question, as written because it's not clear what you mean by "BE and DE is bisector"? Also, what have you tried to solve this problem? $\endgroup$ – B. Mehta Feb 28 '17 at 17:01
  • $\begingroup$ DE is bisector of ADF? $\endgroup$ – Lozenges Feb 28 '17 at 17:56
  • $\begingroup$ Sorry, i edited the question. $\endgroup$ – user373239 Feb 28 '17 at 17:59
  • $\begingroup$ Maybe you should edit your picture instead. $\endgroup$ – Mick Feb 28 '17 at 18:05
  • $\begingroup$ Which part i should edit? $\endgroup$ – user373239 Feb 28 '17 at 18:06

Given triangle $ABD$ and the excircle opposite angle $B$

Given $BF$ tangent to the excircle at $F$ and $BF=15$

$BA$ extended is tangent at $G$ and $BG=BF=15$. Therefore $AG=15-x$

$AD$ is tangent at $M$ with $AM=15-x$, $DM=5$, $AD=20-x=8$


Note: $x=AB$


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