Question 45
We consider $D$, the midpoint of the base $[BC]$ of the isosceles triangle $ABC$, in which $m(\angle BAC) \lt 90^\circ$. On the perpendicular line in $B$ on the line $BC$, one takes the point $E$, such that $\angle EAB \equiv \angle BAC$, and on the parallel through $C$ to the line $AB$, one takes the point $F$, such that $F$ and $D$ are on different sides of the line $AC$ and $\angle FAC \equiv \angle CAD$. Prove that $AE = CF$ and $BF = EF$.
Sorin Furtună
Sorin Furtună