Subfield: Geometry


Question: Let $A B C D$ be a convex quadrilateral and let $P$ and $Q$ be points in $A B C D$ such that $P Q D A$ and $Q P B C$ are cyclic quadrilaterals. Suppose that there exists a point $E$ on the line segment $P Q$ such that $\angle P A E=\angle Q D E$ and $\angle P B E=\angle Q C E$. Show that the quadrilateral $A B C D$ is cyclic.

Solution:
Let $F$ be the point on the line $A D$ such that $E F \| P A$. By hypothesis, the quadrilateral $P Q D A$ is cyclic. So if $F$ lies between $A$ and $D$ then $\angle E F D=\angle P A D=180^{\circ}-\angle E Q D$; the points $F$ and $Q$ are on distinct sides of the line $D E$ and we infer that $E F D Q$ is a cyclic quadrilateral. And if $D$ lies between $A$ and $F$ then a similar argument shows that $\angle E F D=\angle E Q D$; but now the points $F$ and $Q$ lie on the same side of $D E$, so that $E D F Q$ is a cyclic quadrilateral.

In either case we obtain the equality $\angle E F Q=\angle E D Q=\angle P A E$ which implies that $F Q \| A E$. So the triangles $E F Q$ and $P A E$ are either homothetic or parallel-congruent. More specifically, triangle $E F Q$ is the image of $P A E$ under the mapping $f$ which carries the points $P, E$ respectively to $E, Q$ and is either a homothety or translation by a vector. Note that $f$ is uniquely determined by these conditions and the position of the points $P, E, Q$ alone.

Let now $G$ be the point on the line $B C$ such that $E G \| P B$. The same reasoning as above applies to points $B, C$ in place of $A, D$, implying that the triangle $E G Q$ is the image of $P B E$ under the same mapping $f$. So $f$ sends the four points $A, P, B, E$ respectively to $F, E, G, Q$.

If $P E \neq Q E$, so that $f$ is a homothety with a centre $X$, then the lines $A F, P E, B G$-i.e. the lines $A D, P Q, B C$-are concurrent at $X$. And since $P Q D A$ and $Q P B C$ are cyclic quadrilaterals, the equalities $X A \cdot X D=X P \cdot X Q=X B \cdot X C$ hold, showing that the quadrilateral $A B C D$ is cyclic.

Finally, if $P E=Q E$, so that $f$ is a translation, then $A D\|P Q\| B C$. Thus $P Q D A$ and $Q P B C$ are isosceles trapezoids. Then also $A B C D$ is an isosceles trapezoid, hence a cyclic quadrilateral.

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Here is another way to reach the conclusion that the lines $A D, B C$ and $P Q$ are either concurrent or parallel. From the cyclic quadrilateral $P Q D A$ we get

$$
\angle P A D=180^{\circ}-\angle P Q D=\angle Q D E+\angle Q E D=\angle P A E+\angle Q E D .
$$



Hence $\angle Q E D=\angle P A D-\angle P A E=\angle E A D$. This in view of the tangent-chord theorem means that the circumcircle of triangle $E A D$ is tangent to the line $P Q$ at $E$. Analogously, the circumcircle of triangle $E B C$ is tangent to $P Q$ at $E$.

Suppose that the line $A D$ intersects $P Q$ at $X$. Since $X E$ is tangent to the circle $(E A D)$, $X E^{2}=X A \cdot X D$. Also, $X A \cdot X D=X P \cdot X Q$ because $P, Q, D, A$ lie on a circle. Therefore $X E^{2}=X P \cdot X Q$.

It is not hard to see that this equation determines the position of the point $X$ on the line $P Q$ uniquely. Thus, if $B C$ also cuts $P Q$, say at $Y$, then the analogous equation for $Y$ yields $X=Y$, meaning that the three lines indeed concur. In this case, as well as in the case where $A D\|P Q\| B C$, the concluding argument is the same as in the first solution.

It remains to eliminate the possibility that e.g. $A D$ meets $P Q$ at $X$ while $B C \| P Q$. Indeed, $Q P B C$ would then be an isosceles trapezoid and the angle equality $\angle P B E=\angle Q C E$ would force that $E$ is the midpoint of $P Q$. So the length of $X E$, which is the geometric mean of the lengths of $X P$ and $X Q$, should also be their arithmetic mean -impossible, as $X P \neq X Q$. The proof is now complete.