Subfield: Geometry


Question: In the diagram, line segments $A C$ and $D F$ are tangent to the circle at $B$ and $E$, respectively. Also, $A F$ intersects the circle at $P$ and $R$, and intersects $B E$ at $Q$, as shown. If $\angle C A F=35^{\circ}, \angle D F A=30^{\circ}$, and $\angle F P E=25^{\circ}$, determine the measure of $\angle P E Q$.

![](https://cdn.mathpix.com/cropped/2023_12_21_e1c85d542a446b534fb3g-1.jpg?height=602&width=550&top_left_y=924&top_left_x=1232)

Solution:
Let $\angle P E Q=\theta$.

Join $P$ to $B$.

We use the fact that the angle between a tangent to a circle and a chord in that circle that passes through the point of tangency equals the angle inscribed by that chord. We prove this fact below.

More concretely, $\angle D E P=\angle P B E$ (using the chord $E P$ and the tangent through $E$ ) and $\angle A B P=\angle P E Q=\theta$ (using the chord $B P$ and the tangent through $B$ ).

Now $\angle D E P$ is exterior to $\triangle F E P$ and so $\angle D E P=\angle F P E+\angle E F P=25^{\circ}+30^{\circ}$, and so $\angle P B E=\angle D E P=55^{\circ}$.

Furthermore, $\angle A Q B$ is an exterior angle of $\triangle P Q E$.

Thus, $\angle A Q B=\angle Q P E+\angle P E Q=25^{\circ}+\theta$.

![](https://cdn.mathpix.com/cropped/2023_12_21_47eae7df5f8f4c5855a9g-1.jpg?height=604&width=531&top_left_y=744&top_left_x=900)

In $\triangle A B Q$, we have $\angle B A Q=35^{\circ}, \angle A B Q=\theta+55^{\circ}$, and $\angle A Q B=25^{\circ}+\theta$.

Thus, $35^{\circ}+\left(\theta+55^{\circ}\right)+\left(25^{\circ}+\theta\right)=180^{\circ}$ or $115^{\circ}+2 \theta=180^{\circ}$, and so $2 \theta=65^{\circ}$.

Therefore $\angle P E Q=\theta=\frac{1}{2}\left(65^{\circ}\right)=32.5^{\circ}$.

As an addendum, we prove that the angle between a tangent to a circle and a chord in that circle that passes through the point of tangency equals the angle inscribed by that chord.

Consider a circle with centre $O$ and a chord $X Y$, with tangent $Z X$ meeting the circle at $X$. We prove that if $Z X$ is tangent to the circle, then $\angle Z X Y$ equals $\angle X W Y$ whenever $W$ is a point on the circle on the opposite side of $X Y$ as $X Z$ (that is, the angle subtended by $X Y$ on the opposite side of the circle).

We prove this in the case that $\angle Z X Y$ is acute. The cases where $\angle Z X Y$ is a right angle or an obtuse angle are similar.

Draw diameter $X O V$ and join $V Y$.

![](https://cdn.mathpix.com/cropped/2023_12_21_47eae7df5f8f4c5855a9g-1.jpg?height=409&width=374&top_left_y=1720&top_left_x=1583)

Since $\angle Z X Y$ is acute, points $V$ and $W$ are on the same arc of chord $X Y$.

This means that $\angle X V Y=\angle X W Y$, since they are angles subtended by the same chord.

Since $O X$ is a radius and $X Z$ is a tangent, then $\angle O X Z=90^{\circ}$.

Thus, $\angle O X Y+\angle Z X Y=90^{\circ}$.

Since $X V$ is a diameter, then $\angle X Y V=90^{\circ}$.

From $\triangle X Y V$, we see that $\angle X V Y+\angle V X Y=90^{\circ}$.

But $\angle O X Y+\angle Z X Y=90^{\circ}$ and $\angle X V Y+\angle V X Y=90^{\circ}$ and $\angle O X Y=\angle V X Y$ tells us that $\angle Z X Y=\angle X V Y$.

This gives us that $\angle Z X Y=\angle X W Y$, as required.

Final answer: $32.5^{\circ}$