Let $\vartriangle ABC$ be an isosceles triangle with $|AB| = |AC|$. Let $D, E$ and $F$ be points on line segments $BC, CA$ and $AB$, respectively, such that $|BF| = |BE|$ and such that $ED$ is the internal angle bisector of $\angle BEC$. Prove that $|BD|= |EF|$ if and only if $|AF| = |EC|$.
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Tags: geometry, equal segments, isosceles, angle bisector
george_54
05.08.2019 10:12
parmenides51 wrote: Let $\vartriangle ABC$ be an isosceles triangle with $|AB| = |AC|$. Let $D, E$ and $F$ be points on line segments $BC, CA$ and $AB$, respectively, such that $|BF| = |BE|$ and such that $ED$ is the exterior angle bisector of $\angle BEC$. Prove that $|BD|= |EF|$ if and only if $|AF| = |EC|$. 2016 Dutch IMO TST2 P3 May be it is internal?
parmenides51
05.08.2019 13:57
yes there was a typo in their English translation, in the original Dutch it was internal instead of external
george_54
05.08.2019 17:36
$\frac{{BD}}{{DC}} = \frac{{BE}}{{EC}} = \frac{{BF}}{{EC}} \Leftrightarrow \frac{{BD}}{{BF}} = \frac{{DC}}{{EC}}$ and because of $\angle B=\angle C,$ triangles $FBD, ECD$ are similar and $BFED$ cyclic. i) If $AF=EC,$ then $EA = BE \Leftrightarrow \angle A = \theta \Leftrightarrow DE||AB.$ Hence, $BFED$ ia isosceles trapezoid and $\boxed{BD=EF}$ ii) If $BD=EF,$ then $BFED$ ia isosceles trapezoid and $DE||AB$ ...etc
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