Subfield: Geometry


Question: In the diagram, triangle $A B C$ has a right angle at $B$ and $M$ is the midpoint of $B C$. A circle is drawn using $B C$ as its diameter. $P$ is the point of intersection of the circle with $A C$. The tangent to the circle at $P$ cuts $A B$ at $Q$. Prove that $Q M$ is parallel to $A C$.

![](https://cdn.mathpix.com/cropped/2023_12_21_8c2daaa43e7e62803ac0g-1.jpg?height=290&width=414&top_left_y=1514&top_left_x=1365)

Solution:
Since $M$ is the midpoint of a diameter of the circle, $M$ is the centre of the circle.

Join $P$ to $M$. Since $Q P$ is tangent to the circle, $P M$ is perpendicular to $Q P$.

Since $P M$ and $B M$ are both radii of the circle, then $P M=M B$.

![](https://cdn.mathpix.com/cropped/2023_12_21_d64fac5709fb52027973g-1.jpg?height=338&width=468&top_left_y=1300&top_left_x=1357)

Therefore, $\triangle Q P M$ and $\triangle Q B M$ are congruent (Hypotenuse - Side).

Thus, let $\angle M Q B=\angle M Q P=\theta$. So $\angle Q M B=\angle Q M P=90^{\circ}-\theta$

Then $\angle P M C=180^{\circ}-\angle P M Q-\angle B M Q=180^{\circ}-\left(90^{\circ}-\theta\right)-\left(90^{\circ}-\theta\right)=2 \theta$.

But $\triangle P M C$ is isosceles with $P M=M C$ since $P M$ and $M C$ are both radii.

Therefore, $\angle C P M=\frac{1}{2}\left(180^{\circ}-\angle P M C\right)=90^{\circ}-\theta$.

But then $\angle C P M=\angle P M Q$, and since $P M$ is a transversal between $A C$ and $Q M$, then $Q M$ is parallel to $A C$ because of equal alternating angles.
Join $M$ to $P$ and $B$ to $P$.

Since $Q P$ and $Q B$ are tangents to the circle coming from the same point, they have the same length. Since $Q M$ joins the point of intersection of the tangents to the centre of the circle, then by symmetry, $\angle P Q M=\angle B Q M$ and $\angle P M Q=\angle B M Q$. So let $\angle P Q M=\angle B Q M=x$ and $\angle P M Q=\angle B M Q=y$.

![](https://cdn.mathpix.com/cropped/2023_12_21_07eb8e1abd34a8bdc722g-1.jpg?height=341&width=477&top_left_y=301&top_left_x=1298)

Looking at $\triangle Q M B$, we see that $x+y=90^{\circ}$, since $\triangle Q M B$ is right-angled.

Now if we consider the chord $P B$, we see that its central angle is $2 y$, so any angle that it subtends on the circle (eg. $\angle P C B$ ) is equal to $y$.

Thus, $\angle A C B=\angle Q M B$, so $Q M$ is parallel to $A C$.
Join $P B$.

Since $Q P$ is tangent to the circle, then by the Tangent-Chord Theorem, $\angle Q P B=\angle P C B=x$ (ie. the inscribed angle of a chord is equal to the angle between the tangent and chord.

![](https://cdn.mathpix.com/cropped/2023_12_21_07eb8e1abd34a8bdc722g-1.jpg?height=338&width=482&top_left_y=1029&top_left_x=1296)

Since $B C$ is a diameter of the circle, then $\angle C P B=90^{\circ}$ and so $\angle A P B=90^{\circ}$, whence

$\angle A P Q=90^{\circ}-\angle Q P B=90^{\circ}-x$.

Looking at $\triangle A B C$, we see that $\angle P A Q=90^{\circ}-x$, so $\angle P A Q=\angle A P Q$, and so $A Q=Q P$.

But $Q P$ and $Q B$ are both tangents to the circle $(Q B$ is tangent since it is perpendicular to a radius), so $Q P=Q B$.

But then $A Q=Q B$ and $B M=M C$, so $Q$ is the midpoint of $A B$ and $M$ is the midpoint of $B C$. Thus we can conclude that $Q M$ is parallel to $A C$.

(To justify this last statement, we can show very easily that $\triangle Q B M$ is similar to $\triangle A B C$, and so show that $\angle C A B=\angle M Q B$.)

Final answer: 证明题，略