Subfield: Geometry


Question: Given a cyclic quadrilateral $A B C D$, let the diagonals $A C$ and $B D$ meet at $E$ and the lines $A D$ and $B C$ meet at $F$. The midpoints of $A B$ and $C D$ are $G$ and $H$, respectively. Show that $E F$ is tangent at $E$ to the circle through the points $E, G$, and $H$.

Solution:
It suffices to show that $\angle H E F=\angle H G E$ (see Figure 1), since in circle $E G H$ the angle over the chord $E H$ at $G$ equals the angle between the tangent at $E$ and $E H$.

First, $\angle B A D=180^{\circ}-\angle D C B=\angle F C D$. Since triangles $F A B$ and $F C D$ have also a common interior angle at $F$, they are similar.

<img_3775>

Figure 1

Denote by $\mathcal{T}$ the transformation consisting of a reflection at the bisector of $\angle D F C$ followed by a dilation with center $F$ and factor of $\frac{F A}{F C}$. Then $\mathcal{T}$ maps $F$ to $F, C$ to $A, D$ to $B$, and $H$ to $G$. To see this, note that $\triangle F C A \sim \triangle F D B$, so $\frac{F A}{F C}=\frac{F B}{F D}$. Moreover, as $\angle A D B=\angle A C B$, the image of the line $D E$ under $\mathcal{T}$ is parallel to $A C$ (and passes through $B$ ) and similarly the image of $C E$ is parallel to $D B$ and passes through $A$. Hence $E$ is mapped to the point $X$ which is the fourth vertex of the parallelogram $B E A X$. Thus, in particular $\angle H E F=\angle F X G$.

As $G$ is the midpoint of the diagonal $A B$ of the parallelogram $B E A X$, it is also the midpoint of $E X$. In particular, $E, G, X$ are collinear, and $E X=2 \cdot E G$.

Denote by $Y$ the fourth vertex of the parallelogram $D E C Y$. By an analogous reasoning as before, it follows that $\mathcal{T}$ maps $Y$ to $E$, thus $E, H, Y$ are collinear with $E Y=2 \cdot E H$. Therefore, by the intercept theorem, $H G \| X Y$.

From the construction of $\mathcal{T}$ it is clear that the lines $F X$ and $F E$ are symmetric with respect to the bisector of $\angle D F C$, as are $F Y$ and $F E$. Thus, $F, X, Y$ are collinear, which together with $H G \| X Y$ implies $\angle F X E=\angle H G E$. This completes the proof.
We use the following

Lemma (Gauß). Let $A B C D$ be a quadrilateral. Let $A B$ and $C D$ intersect at $P$, and $B C$ and $D A$ intersect at $Q$. Then the midpoints $K, L, M$ of $A C, B D$, and $P Q$, respectively, are collinear.

Proof: Let us consider the points $Z$ that fulfill the equation

$$
(A B Z)+(C D Z)=(B C Z)+(D A Z),
\tag{1}
$$

where $(R S T)$ denotes the oriented area of the triangle $R S T$ (see Figure 2).

<img_3917>

Figure 2

As (1) is linear in $Z$, it can either characterize a line, or be contradictory, or be trivially fulfilled for all $Z$ in the plane. If (1) was fulfilled for all $Z$, then it would hold for $Z=A, Z=B$, which gives $(C D A)=(B C A),(C D B)=(D A B)$, respectively, i.e. the diagonals of $A B C D$ would bisect each other, thus $A B C D$ would be a parallelogram. This contradicts the hypothesis that $A D$ and $B C$ intersect. Since $E, F, G$ fulfill (1), it is the equation of a line which completes the proof of the lemma.

Now consider the parallelograms $E A X B$ and $E C Y D$ (see Figure 1). Then $G, H$ are the midpoints of $E X, E Y$, respectively. Let $M$ be the midpoint of $E F$. By applying the Lemma to the (re-entrant) quadrilateral $A D B C$, it is evident that $G, H$, and $M$ are collinear. A dilation by a factor of 2 with center $E$ shows that $X, Y, F$ are collinear. Since $A X \| D E$ and $B X \| C E$, we have pairwise equal interior angles in the quadrilaterals $F D E C$ and $F B X A$. Since we have also $\angle E B A=\angle D C A=\angle C D Y$, the quadrilaterals are similar. Thus, $\angle F X A=\angle C E F$.

Clearly the parallelograms $E C Y D$ and $E B X A$ are similar, too, thus $\angle E X A=\angle C E Y$. Consequently, $\angle F X E=\angle F X A-\angle E X A=\angle C E F-\angle C E Y=\angle Y E F$. By the converse of the tangent-chord angle theorem $E F$ is tangent to the circle $X E Y$. A dilation by a factor of $\frac{1}{2}$ completes the proof.
As in Solution 2, G, H, M are proven to be collinear. It suffices to show that $M E^{2}=M G \cdot M H$. If $\boldsymbol{p}=\overrightarrow{O P}$ denotes the vector from circumcenter $O$ to point $P$, the claim becomes

$$
\left(\frac{\boldsymbol{e}-\boldsymbol{f}}{2}\right)^{2}=\left(\frac{\boldsymbol{e}+\boldsymbol{f}}{2}-\frac{\boldsymbol{a}+\boldsymbol{b}}{2}\right)\left(\frac{\boldsymbol{e}+\boldsymbol{f}}{2}-\frac{\boldsymbol{c}+\boldsymbol{d}}{2}\right)
$$

or equivalently

$$
4 \boldsymbol{e} \boldsymbol{f}-(\boldsymbol{e}+\boldsymbol{f})(\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}+\boldsymbol{d})+(\boldsymbol{a}+\boldsymbol{b})(\boldsymbol{c}+\boldsymbol{d})=0
\tag{2}
$$

With $R$ as the circumradius of $A B C D$, we obtain for the powers $\mathcal{P}(E)$ and $\mathcal{P}(F)$ of $E$ and $F$, respectively, with respect to the circumcircle

$$
\begin{aligned}
& \mathcal{P}(E)=(\boldsymbol{e}-\boldsymbol{a})(\boldsymbol{e}-\boldsymbol{c})=(\boldsymbol{e}-\boldsymbol{b})(\boldsymbol{e}-\boldsymbol{d})=\boldsymbol{e}^{2}-R^{2} \\
& \mathcal{P}(F)=(\boldsymbol{f}-\boldsymbol{a})(\boldsymbol{f}-\boldsymbol{d})=(\boldsymbol{f}-\boldsymbol{b})(\boldsymbol{f}-\boldsymbol{c})=\boldsymbol{f}^{2}-R^{2}
\end{aligned}
$$

hence

$$
(\boldsymbol{e}-\boldsymbol{a})(\boldsymbol{e}-\boldsymbol{c})=\boldsymbol{e}^{2}-R^{2}
\tag{3}
$$
$$
(\boldsymbol{e}-\boldsymbol{b})(\boldsymbol{e}-\boldsymbol{d})=\boldsymbol{e}^{2}-R^{2}
\tag{4}
$$
$$
(\boldsymbol{f}-\boldsymbol{a})(\boldsymbol{f}-\boldsymbol{d})=\boldsymbol{f}^{2}-R^{2}
\tag{5}
$$
$$
(\boldsymbol{f}-\boldsymbol{b})(\boldsymbol{f}-\boldsymbol{c})=\boldsymbol{f}^{2}-R^{2}
\tag{6}
$$

Since $F$ lies on the polar to $E$ with respect to the circumcircle, we have

$$
4 \boldsymbol{e} \boldsymbol{f}=4 R^{2}
\tag{7}
$$

Adding up (3) to (7) yields (2), as desired.