<s>[INST] Two sides of quadrilateral $ABCD$ intersect at $E$ Quadrilateral $ABCD$ is inscribed in a circle. The extensions of $AD$ and $BC$ intersect at $E$, such that $D$ is between $A$ and $E$. If $BC=2DE,AD:EC=7:2$ and $\cos\measuredangle AEB=\dfrac78$, find $\cos\measuredangle ABC$.

Since $\measuredangle ABC+\measuredangle ADC=180^\circ,$ angle $\measuredangle CDE$ is also equal to $\beta$. Triangles $ABE$ and $CDE$ also share $\measuredangle E$. By AA Similarity, $\triangle ABE\sim\triangle CDE$ with a ratio of $\dfrac{AB}{CD}=\dfrac{BE}{DE}=\dfrac{AE}{CE}$.
$Let DE=x, BC=2x$ and $AD=7y, EC=2y$. We have to find $$\cos\beta=\dfrac{AB^2+BE^2-AE^2}{2\cdot AB\cdot BE}$$ and we have $$\cos\varphi=\dfrac{ED^2+EC^2-CD^2}{2\cdot ED\cdot EC}=\dfrac{x^2+4y^2-CD^2}{2\cdot x\cdot 2y}=\dfrac{x^2+4y^2-CD^2}{4xy}$$ I don't see how to relate $AB$ and $CD$ (other than the similar triangles) Thank you in advance!
 [/INST] Hint:
From the similarity of $\triangle ABE$ and $\triangle CDE$ we have
$$\frac{x}{2y} = \frac{2x+2y}{x+7y}$$
$$x^2+7xy = 4xy + 4y^2 $$
$$(x+4y)(x-y) = 0$$
Can you take it from here?
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