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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.

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    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? – 2017-02-28
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    DE is bisector of ADF? – 2017-02-28
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    Sorry, i edited the question. – 2017-02-28
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    Maybe you should edit your picture instead. – 2017-02-28
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    Which part i should edit? – 2017-02-28
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    Enlarge it so that the given can be seen more clearly. – 2017-02-28
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    I think it is visible when clicked. – 2017-02-28
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    Didn't realize your picture can be magnified after click. Also, the necessary data have also been included in the edited version. – 2017-03-01

1 Answers 1

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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 $AM=15-x$, $DM=5$, $AD=20-x=8$

$x=12$

Note: $x=AB$