Subfield: Geometry


Question: Let $A B C$ be an acute-angled triangle with circumcircle $\Gamma$. Let $D$ and $E$ be points on the segments $A B$ and $A C$, respectively, such that $A D=A E$. The perpendicular bisectors of the segments $B D$ and $C E$ intersect the small arcs $\overparen{A B}$ and $\overparen{A C}$ at points $F$ and $G$ respectively. Prove that $D E \| F G$.

Solution:
In the sequel, all the considered arcs are small arcs.

Let $P$ be the midpoint of the arc $\overparen{B C}$. Then $A P$ is the bisector of $\angle B A C$, hence, in the isosceles triangle $A D E, A P \perp D E$. So, the statement of the problem is equivalent to $A P \perp F G$.

In order to prove this, let $K$ be the second intersection of $\Gamma$ with $F D$. Then the triangle $F B D$ is isosceles, therefore

$$
\angle A K F=\angle A B F=\angle F D B=\angle A D K,
$$

yielding $A K=A D$. In the same way, denoting by $L$ the second intersection of $\Gamma$ with $G E$, we get $A L=A E$. This shows that $A K=A L$.

<img_3468>

Now $\angle F B D=\angle F D B$ gives $\overparen{A F}=\overparen{B F}+\overparen{A K}=\overparen{B F}+\overparen{A L}$, hence $\overparen{B F}=\overparen{L F}$. In a similar way, we get $\overparen{C G}=\overparen{G K}$. This yields

$$
\angle(A P, F G)=\frac{\overparen{A F}+\overparen{P G}}{2}=\frac{\overparen{A L}+\overparen{L F}+\overparen{P C}+\overparen{C G}}{2}=\frac{\overparen{K L}+\overparen{L B}+\overparen{B C}+\overparen{C K}}{4}=90^{\circ}
$$
Let $Z=A B \cap F G, T=A C \cap F G$. It suffices to prove that $\angle A T Z=\angle A Z T$.

Let $X$ be the point for which $F X A D$ is a parallelogram. Then

$$
\angle F X A=\angle F D A=180^{\circ}-\angle F D B=180^{\circ}-\angle F B D \text {, }
$$

where in the last equality we used that $F D=F B$. It follows that the quadrilateral $B F X A$ is cyclic, so $X$ lies on $\Gamma$.

<img_4011>



Analogously, if $Y$ is the point for which $G Y A E$ is a parallelogram, then $Y$ lies on $\Gamma$. So the quadrilateral $X F G Y$ is cyclic and $F X=A D=A E=G Y$, hence $X F G Y$ is an isosceles trapezoid.

Now, by $X F \| A Z$ and $Y G \| A T$, it follows that $\angle A T Z=\angle Y G F=\angle X F G=\angle A Z T$.
As in the first solution, we prove that $F G \perp A P$, where $P$ is the midpoint of the small arc $\overparen{B C}$.

Let $O$ be the circumcentre of the triangle $A B C$, and let $M$ and $N$ be the midpoints of the small arcs $\overparen{A B}$ and $\overparen{A C}$, respectively. Then $O M$ and $O N$ are the perpendicular bisectors of $A B$ and $A C$, respectively.

<img_3153>

The distance $d$ between $O M$ and the perpendicular bisector of $B D$ is $\frac{1}{2} A B-\frac{1}{2} B D=\frac{1}{2} A D$, hence it is equal to the distance between $O N$ and the perpendicular bisector of $C E$.

This shows that the isosceles trapezoid determined by the diameter $\delta$ of $\Gamma$ through $M$ and the chord parallel to $\delta$ through $F$ is congruent to the isosceles trapezoid determined by the diameter $\delta^{\prime}$ of $\Gamma$ through $N$ and the chord parallel to $\delta^{\prime}$ through $G$. Therefore $M F=N G$, yielding $M N \| F G$.

Now

$$
\angle(M N, A P)=\frac{1}{2}(\overparen{A M}+\overparen{P C}+\overparen{C N})=\frac{1}{4}(\overparen{A B}+\overparen{B C}+\overparen{C A})=90^{\circ}
$$

hence $M N \perp A P$, and the conclusion follows.