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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 $\overset{\LARGE{\frown}}{QF}=72^o$ enter image description here

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...???

enter image description here

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2 Answers 2

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

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  • $\begingroup$ One question..Shouldn't $PL$ be tangent to $\angle PLM=36^o$? $\endgroup$ Commented Jan 28, 2022 at 9:52
  • $\begingroup$ @petaarantes no, do you see that $\angle FLQ = 144^0$? $\endgroup$
    – Math Lover
    Commented Jan 28, 2022 at 10:08
  • $\begingroup$ What is this property of the angle opposite the central being double in the quadrilateral $OFLQ$? $\endgroup$ Commented Jan 28, 2022 at 10:23
  • $\begingroup$ What is the angle arc $FQ$ subtends at circumference opposite $L$? $~36^\circ$, right? So, $\angle FLQ = 180^0 - 36^0$. $\endgroup$
    – Math Lover
    Commented Jan 28, 2022 at 10:27
  • $\begingroup$ 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 $\endgroup$ Commented Jan 28, 2022 at 10:53
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enter image description here

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:

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

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  • $\begingroup$ You are welcome. $\endgroup$
    – sirous
    Commented Jan 28, 2022 at 13:01

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