Question 6
On the side $[CD]$ of the square $ABCD$ point $E$ is chosen such that $m(\angle ABE) = 60^{\circ}$. On the halfline $(BA$ point $F$ is chosen such that $[BE] = [BF]$. Let $M$ be the intersection between the lines $EF$ and $AD$.
a) Prove that $m(\angle BME) = 75^{\circ}$.
b) The bisector of angle $CBE$ intersects $CD$ at $N$. Prove that the triangle $BMN$ is equilateral.
a) Prove that $m(\angle BME) = 75^{\circ}$.
b) The bisector of angle $CBE$ intersects $CD$ at $N$. Prove that the triangle $BMN$ is equilateral.