Marinada '23

[lang = hr] U paralelogramu $ABCD (AB \parallel CD, AD \parallel BC)$ vrijedi $AD=BD$. Neka je $E$ točka na dužini $BD$ tako da vrijedi $AE=DE$. Pravac $AE$ siječe dužinu $BC$ u točki $F$. Ispostavilo se da je pravac $DF$ simetrala $\angle CDE$. Izračunaj $\angle ABD$. Rješenje zaokruži na dvije decimale. [/lang] [lang = en] In the parallelogram $ABCD (AB \parallel CD, AD \parallel BC)$ $AD=BD$ holds. Let $E$ be a point on the length $BD$ such that $AE=DE$ holds. Line $AE$ intersects length $BC$ at point $F$. It turns out that the line $DF$ is the bisector of $\angle CDE$. Calculate $\angle ABD$. Round the solution to two decimal places. [/lang]