Find the $\angle FPA$ in the figure below

Question

For reference: Starting from a point $P$, outside a circle the tangent $PA$ and the secant $PQL$ are drawn. Then join $L$ with the midpoint $M$ of $PA$. LM intersects at F the circle. Calculate $\angle FPA$ if $\overset{\LARGE{\frown}}{QF}=72^o$

My progress:

$\angle FAP = \theta=\angle ALM$ (alternate angles)

$\triangle AOF$(isosceles) $\implies \angle OAF = \angle AFO=90-\theta$

$\angle AOF = 2\theta$

I'm not seeing the other relationships...???

Answer 1

Note that $\angle PLM = 36^\circ$

Also using power of point $M$, $~MA^2 = MF \cdot ML = PM^2$

$$ \implies \frac{PM}{FM} = \frac{ML}{PM}$$

and given $\angle PML$ is common, $$\triangle PLM \sim \triangle FPM~~ \text {(by S-A-S rule)}$$

That leads to $~\angle FPM = \angle PLM = 36^\circ$

Answer 2

In figure circle S is circumcircle of triangle APF. AQ is perpendicular on radius AS.In circle S angle APF is opposite to arc AF, the measure of arc AF is $72^o$ because it is opposite to angle QAF and we have:

$\angle QAF=\frac {72}2=36^o \Rightarrow \overset{\large\frown}{AF}=72^o$

and QA is tangent on circle S at vertex A.

So the measure of angle APF is:

$\angle APF= \frac{72}2=36^o$