Prove two angles on tangents of two circles are equal

Question

Question:

ABCD is a cyclic quadrilateral.

Circle $DCJ(P)$ is tangent to $AB$ at $J$.

Circle $ABK(O)$ is tangent to $DC$ at $K$.

Prove $\angle BJK=\angle CKJ$.

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Attempt

Extend $AB$ and $DC$ so that they meet at point $E$, then the question is equivalent to prove $EJ =EK$. Now, by the Power of a Point Theorem, we know the following: $$EA\cdot EB= EK^2\\ EC\cdot ED=EJ^2 $$

I saw PPL: Apollonius' Problem with Two Points and a Line but I don't know how to use here.

Note: Please share your ideas and your attempts in comments or even answer if your solution is not complete. Thanks!

Answer 1

In the original post is only missing the equality $$EA\cdot EB=EC\cdot ED\ ,$$ which follows from $ABCD$ being inscribed in a circle.

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Picture and solution for the problem, putting all together:

Let $E$ be the point of intersection $AB\cap CD$. Then we have $$ \begin{aligned} EJ^2 &= EC\cdot ED &&\text{(Power of $E$ w.r.t circle $(JCD)$)} \\ &= EB\cdot EA &&\text{(Power of $E$ w.r.t circle $(ABCD)$)} \\ &=EK^2 &&\text{(Power of $E$ w.r.t circle $(ABK)$)} \end{aligned} $$ so $EJ=EK$, the triangle $\Delta EJK$ is isosceles in $E$, so its angles in $J,K$ are equal.

$\square$

Answer 2

Let $JK\cap arc DC=\{N\}$.

Thus, $\measuredangle KJB=\measuredangle JDN$ and if $\measuredangle KJB=\measuredangle JKC$, we obtain $$\measuredangle JDN=\measuredangle DJN+\measuredangle JDK,$$ which gives $$\measuredangle NDC=\measuredangle DJN,$$ which says that $N$ is a mid-point of the arc $DC$, which is not necessary if not given that $ABCD$ is cyclic.