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Geometry Difficulty 6.4 National Olympiad Prove it Taiwan

設 $ABCD$ 為凸四邊形，其中 $∠ABC > 90°$ 、 $∠CDA > 90°$ ，並且 $∠DAB = ∠BCD$ 。令點 $A$ 分別關於直線 $BC$ 與 $CD$ 的對稱點為點 $E$ 與 $F$ 。設線段 $AE$ 和 $AF$ 分別交直線 $BD$ 於點 $K$ 和 $L$ 。試證：三角形 $BEK$ 和 $DFL$ 的兩個外接圓相切。

Let $ABCD$ be a convex quadrilateral with $∠ABC > 90°$ , $∠CDA > 90°$ , and $∠DAB = ∠BCD$ . Denote by $E$ and $F$ the reflections of $A$ in lines $BC$ and $CD$ , respectively. Suppose that the segments $AE$ and $AF$ meet the line $BD$ at $K$ and $L$ , respectively. Prove that the circumcircles of triangles $BEK$ and $DFL$ are tangent to each other.

Solution

Let $A'$ be the reflection of $A$ in $BD$ . In the following we will prove that: the quadrilaterals $A'BKE$ and $A'DLF$ each have a circumcircle, and their circumcircles are tangent to each other at $A'$ .

By reflection in the line $BC$ , we have $∠BEK = ∠BAK$ ; and by reflection in the line $DB$ , we have $∠BAK = ∠BA'K$ . Hence $∠BEK = ∠BA'K$ , which shows that $A'BEK$ is concyclic. Similarly, we obtain that $A'DLF$ is concyclic.

To prove that the circles $A'BKE$ and $A'DFL$ are tangent, it suffices to show that
$∠A'KB + ∠A'LD = ∠BA'D$
Since $AK ⊥ BC$ , $AL ⊥ CD$ , and using once more the reflection in the line $BD$ , we know that
$∠A'KB + ∠A'LD = 180° - ∠KA'L = 180° - ∠KAL = ∠BCD = ∠BAD = ∠BA'D,$
and thus the result is proved.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.