Subfield: Geometry


Question: Let $A B C D$ be an isosceles trapezoid with $A B \| C D$. Let $E$ be the midpoint of $A C$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $A B E$ and $C D E$, respectively. Let $P$ be the crossing poitn of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $P E$ is tangent to $\Omega$.

Solution:
If $A B C D$ is a rectangle, the statement is trivial due to symmetry. Hence, in what follows we assume $A D \nparallel B C$.



Let $F$ be the midpoint of $B D$; by symmetry, both $\omega$ and $\Omega$ pass through $F$. Let $P^{\prime}$ be the meeting point of tangents to $\omega$ at $F$ and to $\Omega$ at $E$. We aim to show that $P^{\prime}=P$, which yields the required result. For that purpose, we show that $P^{\prime} A$ and $P^{\prime} D$ are tangent to $\omega$ and $\Omega$, respectively.



Let $K$ be the midpoint of $A F$. Then $E K$ is a midline in the triangle $A C F$, so $\angle(A E, E K)=$ $\angle(E C, C F)$. Since $P^{\prime} E$ is tangent to $\Omega$, we get $\angle(E C, C F)=\angle\left(P^{\prime} E, E F\right)$. Thus, $\angle(A E, E K)=$ $\angle\left(P^{\prime} E, E F\right)$, so $E P^{\prime}$ is a symmedian in the triangle $A E F$. Therefore, $E P^{\prime}$ and the tangents to $\omega$ at $A$ and $F$ are concurrent, and the concurrency point is $P^{\prime}$ itself. Hence $P^{\prime} A$ is tangent to $\omega$.



The second claim is similar. Taking $L$ to be the midpoint of $D E$, we have $\angle(D F, F L)=$ $\angle(F B, B E)=\angle\left(P^{\prime} F, F E\right)$, so $P^{\prime} F$ is a symmedian in the triangle $D E F$, and hence $P^{\prime}$ is the meeting point of the tangents to $\Omega$ at $D$ and $E$.



<img_3537>
Let $Q$ be the isogonal conjugate of $P$ with respect to $\triangle A E D$, so $\angle(Q A, A D)=$ $\angle(E A, A P)=\angle(E B, B A)$ and $\angle(Q D, D A)=\angle(E D, D P)=\angle(E C, C D)$. Now our aim is to prove that $Q E \| C D$; this will yield that $\angle(E C, C D)=\angle(A E, E Q)=\angle(P E, E D)$, whence $P E$ is tangent to $\Omega$.



Let $D Q$ meet $A B$ at $X$. Then we have $\angle(X D, D A)=\angle(E C, C D)=\angle(E A, A B)$ and $\angle(D A, A X)=\angle(A B, B C)$, hence the triangles $D A X$ and $A B C$ are similar. Since $\angle(A B, B E)=$ $\angle(D A, A Q)$, the points $Q$ and $E$ correspond to each other in these triangles, hence $Q$ is the midpoint of $D X$. This yields that the points $Q$ and $E$ lie on the midline of the trapezoid parallel to $C D$, as desired.







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Let $O$ be the intersection of the diagonals $A C$ and $B D$. Let $F$ be the midpoint of $B D$. Let $S$ be the second intersection point of the circumcircles of triangles $A O F$ and $D O E$. We will prove that $S D$ and $S E$ are tangent to $\Omega$; the symmetric argument would then imply also that $S A$ and $S F$ are tangent to $\Gamma$. Thus $S=P$ and the claimed tangency holds.



We first prove that $O S$ is parallel to $A B$ and $D C$. Compute the powers of the points $A, B$ with respect to the circumcircles of $A O F$ and $D O E$ :



$$

\begin{gathered}

d(A, A O F)=0, \quad d(A, D O E)=A O \cdot A E \\

d(B, A O F)=B O \cdot B F, \quad d(B, D O E)=B O \cdot B D=2 B O \cdot B F

\end{gathered}

$$



And therefore



$$

d(B, D O E)-d(B, A O F)=B O \cdot B F=A O \cdot A E=d(A, D O E)-d(A, A O F)

$$



Thus both $A$ and $B$ belong to a locus of the form



$$

d(X, D O E)-d(X, A O F)=\mathrm{const}

$$



which is always a lines parallel to the radical axis of the respective circles. Since this radical axis is $O S$ by definition, it follows that $A B$ is parallel to $O S$, as claimed.



Now by angle chasing in the cyclic quadrilateral $D S O E$, we find



$$

\begin{aligned}

\angle(S D, D E) & =\angle(S O, O E)=\angle(D C, C E), \\

\angle(S E, E D)=\angle(S O, O D) & =\angle(D C, D B)=\angle(A C, C D)=\angle(E C, C D),

\end{aligned}

$$



and these angle equalities are exactly the conditions of $S D, S E$ being tanget to $\Omega$, as claimed.