For reference: The angle $B$ of an isosceles triangle $ABC (AB = BC)$ measures $48°$. Points $P$ and $Q$ are marked on the $BM$ median. The prolongation of $AQ$ intersects $BC$ in $F$. If $B, P, Q$ and $M$ form a harmonic range, calculate $\angle PFB$.(Answer $66^o$}
My progress:
I made the drawing and the relationship for the segment to be harmonic...but I don't know if there is missing data or there is another relationship with angles because I don't understand much about this subject. by Geogebra $FP$ is angle bisector triangle $BFM$ and $\angle FPB = 90^o$ ...Someone would know how to explain this issue
Condition for the four points $ACBD$ to form a harmonic range: $\frac{BQ}{PQ} = \frac{BM}{MQ}$
BM ís median amd angle bissector therfore $\angle PBF = 24^o$
By solution $\angle PBF =24^0 \therefore FPB = 90^o$
but how did he come to the conclusion that $\angle FPB = 90^o$ ?
As mentioned in the comments above, $\small FP$ need not bisect $\small \angle BFM$. Though the observation $\small BM\perp FP$ holds.
Let's prove the vice versa.
Draw perpendicular line $\small FD$ to $\small BM$.
We see that $\small \triangle BDF\sim\triangle BMC\implies\dfrac{BD}{BM}=\dfrac{DF}{MC}.$
Due to being isosceles, we know $\small MC=AM$.
Again $\small \triangle QDF\sim\triangle QMA\implies\dfrac{DF}{AM}=\dfrac{QD}{QM}.$
Thus combining above, $\small \dfrac{BD}{BM}=\dfrac{QD}{QM}$ or $\small \dfrac{BD}{QD}=\dfrac{BM}{QM}.$
Therefore, $\small B$, $\small D$, $\small Q$, $\small M$ form a harmonic range.
Hence, eventually $\small P\equiv D$.