Isosceles trapezoid and similar triangles

Question

$ABCD$ is an isosceles trapezoid $(AB \parallel CD)$. On the leg $BC$ is chosen point $P$ such that $\angle DAC=\angle CAP$ and on the diagonal $AC$ is chosen point $K$ such that $\dfrac{AD}{AC}=\dfrac{AK}{AP}$. Show that $K$ lies on $DB$.

We have that $\dfrac{AD}{AC}=\dfrac{AK}{AP}$ and $\angle DAK = \angle CAP$. Therefore, $\triangle AKD \sim \triangle APC$ and $\dfrac{AK}{AP}=\dfrac{AD}{AC}=\dfrac{DK}{CP}$. What can I try to show that $K$ lies on $DB$?

Answer 1

Because by your work $$\measuredangle ADK=\measuredangle ACB$$ and $$\measuredangle ACB=\measuredangle ADB.$$ Id est, $$\measuredangle ADK=\measuredangle ADB$$ and we are done!

Answer 2

Let me propose an other solution.

$K \in BD$

$\angle KBA=\angle KAB=\angle BCA $

$\angle CBD=\angle CAP$

$ quadrilateral\;ABPK\; is\; cyclic. Hence,\angle KPA=\angle KBA=\angle DCA $

$\triangle ACD \sim \triangle APK$

$\dfrac{AD}{AC}=\dfrac{AK}{AP}$

Since $\triangle APK$ is unique, point $K$ is the only one corresponding to the given requirements.