# Find the $\angle FPA$ in the figure below

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...???

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 $$~\angle FPM = \angle PLM = 36^\circ$$

• One question..Shouldn't $PL$ be tangent to $\angle PLM=36^o$? Commented Jan 28, 2022 at 9:52
• @petaarantes no, do you see that $\angle FLQ = 144^0$? Commented Jan 28, 2022 at 10:08
• What is this property of the angle opposite the central being double in the quadrilateral $OFLQ$? Commented Jan 28, 2022 at 10:23
• What is the angle arc $FQ$ subtends at circumference opposite $L$? $~36^\circ$, right? So, $\angle FLQ = 180^0 - 36^0$. Commented Jan 28, 2022 at 10:27
• Now I understand.. I had alternates angles in my mind (which was the case of the theta angle) and that's why I was confusing..gratitude Commented Jan 28, 2022 at 10:53

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:

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

• You are welcome. Commented Jan 28, 2022 at 13:01