Lemma: In any triangle $ \triangle ABC$, if $ D$ lies on side side $ BC$, $ \alpha = m \angle BAD$ and $ \beta m \angle DAC$, then $ \frac {AC}{AB} = \frac {DC \sin \beta}{BD \sin \alpha}$. Proof: Let $ \theta = m \angle ADC$. By the law of sines, we have $ \frac {AC}{\sin \theta} = \frac {DC}{\sin \beta} \,\,\,\,\,\,\,\,\,\, (1)$ and $ \frac {AB}{\sin(\pi - \theta)} \& = \nonumber \\ \frac {AB}{\sin \theta} = \frac {BD}{\sin \alpha} \,\,\,\,\,\,\,\,\,\, (2)$ Dividing (1) by (2) yields $ \frac {AC}{AB} = \frac {DC \sin \beta}{BD \sin \alpha}$ as desired. Since $ m \angle ABC = 80^{\circ}$ and $ \triangle ABC$ is isosceles, we have $ m \angle BAC = m \angle BCA = 50^{\circ}$. Let $ \omega$ be the circle tangent to lines $ AB$ and $ BC$, at $ A$ and $ C$. For any point $ X$, we have that $ \angle AXC = 50^{\circ}$ if and only if $ X$ is on the circumference of $ \omega$. Since $ m \angle ADC = 50^{\circ}$, we have that $ D$ lies on $ \omega$. Let $ A_1$ be $ \overline{AC} \cap \overline{BD}$. We want to show that $ m \angle CDE = m \angle ADB$, that is, $ m \angle ADE = m \angle CDA_1$. In other words, we want to show that $ D$ is the point where the reflection of the median through $ D$ to $ AC$ across the angle bisector of $ \angle ADC$, that is, symmedian of $ \triangle ADC$ from $ A$, intersects $ AC$. Let $ Z = BD \cap \omega$. Note that $ \triangle ABZ \sim \triangle DBA$ and $ \triangle CBZ \sim DBC$ by AA (since $ \angle BDA \cong \angle AZB$ and $ \angle CDB \equiv \angle BCZ$.) Thus, $ \frac {BA}{ZA} = \frac {DB}{AD}$ and $ \frac {CB}{CZ} = \frac {DB}{DC}$. Since $ BC = BA$ by equal tangents, we have $ \frac {DB \cdot AZ}{AD} = BA = BC = \frac {DB \cdot CZ}{DC}$ Dividing both sides by $ DB$, we conclude that $ \frac {AZ}{AD} = \frac {CZ}{DC}$, or $ \frac {AZ}{ZC} = \frac {AD}{DC}$. Let $ m \angle ZDA = \alpha$ and $ m \angle CDZ = \beta$. Since $ ABCD$ is cyclic, $ m \angle ZCA = \alpha$ and $ m \angle CAZ = \beta$. By the law of sines, $ \frac {AZ}{\sin \alpha} = \frac {CZ}{\sin \beta}$, so $ \frac {AD}{DC} = \frac {AZ}{ZC} = \frac {\sin \beta}{\sin \alpha}$. However, our lemma tells us that $ \frac {AD}{DC} = \frac {AA_1 \sin \alpha}{CA_1 \sin \beta} = \frac {AA_1}{CA_1} \frac {DC}{AD}$. Hence, $ \frac {AA_1}{CA_1} = \frac {AD^2}{DC^2} \,\,\,\,\,\,\,\,\,\, (3)$ Now, let $ A_1'$ be the point where the symmedian of $ \triangle ACD$ through $ D$ hits $ AC$. Since the function $ f(x) = \frac {x}{a - x} = \frac {a}{a - x} - 1$ is injective for $ x \in (0, s)$, if we can show that $ \frac {AA_1'}{AC - AA_1'} = \frac {AA_1'}{A_1 C} = \frac {AD^2}{DC^2}$, it will follow by (3) that $ A_1 = A_1'$. Let $ \alpha' = m \angle CDE$ and $ \beta' = m \angle EDA$. Our lemma tells us that $ \frac {\sin \beta'}{\sin \alpha'} \& = \frac {AD \cdot CE}{DC \cdot EA} \\ \& = \frac {AD}{DC} \,\,\,\,\,\,\,\,\,\, (4)$ since $ E$ is the midpoint of $ AC$. However, by definition, $ \alpha' = m \angle A_1' DA$ and $ \beta' = m \angle A_1' DC$, so by our lemma and (4), \begin{align*} \frac {AA_1'}{A_1' C} & = \frac {DA \sin \alpha'}{DC \sin \beta'} \\ & = \frac {DA}{DC} \frac {1}{\frac {DC}{AD}} \\ & = \frac {DA^2}{DC^2} \end{align*} which was what we wanted.

Put $ \angle CDE=x,\angle ADB=y,\angle ABD=z$. Use the sine theorem in triangles $ ADE$ and $ DEC$ to obtain: $ \frac{AD}{AE}=\frac{\sin \angle AED}{\sin(50-x)}$ $ \frac{CE}{DC}=\frac{\sin x}{\sin\angle DEC}$ Multiply these two to get ($ AE=CE$ as $ ABC$ is isosceles): $ \frac{AD}{DC}=\frac{\sin x}{\sin 50-x}$ In $ ADC$, sine theorem gives us $ \frac{AD}{DC}=\frac{\sin(y+z)}{\sin(130-y-z)}$ Also, $ 1=\frac{AB}{BD}\cdot\frac{BD}{BC}=$ $ \frac{\sin y}{\sin(180-y-z)}\cdot \frac{\sin(50+y+z)}{\sin(50-y)}=$ $ \frac{\sin y}{\sin(50-y)}\cdot\frac{DC}{AD}$ (because $ \sin(180-t)=\sin t$) Now we get $ \frac{\sin y}{\sin(50-y)}=\frac{\sin x}{\sin(50-x)}$ Cross-mutliply and use product-to-sum formulas to get: $ \cos(50-x-y)-\cos(50-x+y)=\cos(50-y-x)-\cos(50-y+x)$ $ \cos(50-x+y)=\cos(50-y+x)$ The values in brackets vary from $ 0$ to $ 100$, and as $ \cos$ function is strictly decreasing in that interval, the 2 brackets must be equal, so $ x=y$