indeed if CE cuts AB at P, A is the circumcenter of triangle PBF

proglote wrote: indeed if CE cuts AB at P, A is the circumcenter of triangle PBF Nice solution, but you need to finish the proof as such: $PEA$ and $PCB$ are similar triangles since $EA \parallel CB$ and $\angle EPA = \angle CPB$. Since $CB/EA = 2$, $BP/AP = 2$, so we get $AP = AB$. Since $\angle PBF = 90$ and $AP=AB$, we conclude that $A$ is the circumcenter of $PBF$. Thus, $AF=AB$ which means $AFB$ is isosceles. You have to make sure that you're not missing any steps in the proof.

Dear Mathlinkers, https://artofproblemsolving.com/community/c6t48f6h1423260_parallelogram Sincerely Jean-Louis

Let $M$ be a point at the line CD such that distance from $M$ to $C$ is shorter than distance frok $M$ to $D$ and $MC = CD = AB$ then take $P = AM \cap BN$ and it is easy to prove that $AP$ is median ans height at triangle $ABF$.

Let $E'$ be the midpoint of $BC$ and $F'=AE'\cap BF$. By symmetry $\angle ECB=\angle E'AD\Rightarrow\angle ECB=\angle AE'B\Rightarrow AE'\mid\mid CE$ But that means $E'F'$ is midbase of $\Delta BE'F'\Rightarrow BF'=FF'$. Also $\angle BF'A=\angle BFC=90^{\circ}$. Therefore $AF'$ is simultaneously height and median of $\Delta ABF\Rightarrow\Delta ABF$ is isosceles, $Q.E.D.$

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