If I am not wrong, Part a is trivial because \[\angle ABC=\angle DBA+\angle DBC=80\]and since it's isoceles we're done.

For the part (b) we denote by $\alpha$ measure of the $\angle DAC$. By the Law of Sines in $\triangle ABD$ and $\triangle ACD$ we have that $$1 = \frac{BA}{AD}\cdot \frac{DA}{AC} = \frac{\sin (70^\circ - \alpha)}{\sin 50^\circ} \cdot \frac{\sin 25^\circ}{\sin (25^\circ + \alpha)}$$ Claim. $\alpha = 5^\circ$ works Proof. We need to show that $\sin 30^\circ \cdot \sin 50^\circ = \sin 25^\circ \cdot \sin 65^\circ$. Which is clear, since $2\sin 25^\circ \cdot \sin 65^\circ = 2\sin 25^\circ \cos 25^\circ = \sin 50^\circ$. $\square$ Notice that measure of $\alpha$ is define in unique way by the condition of the problem, so we conclude that answer is $\boxed{5^\circ}$

Pretty sure this is the intended solution: Let $E$ be the point on the extension of $BD$ such that $AB=AE$. Thus, $AE=AC$. Then, $\angle ABE=\angle AEB=50$, and $\angle CAE=\angle BAE-\angle ABC=60$. Thus, $AEC$ is an equilateral triangle and so $AE=EC$. At the same time, $\angle AED$ is twice of $\angle ACD$, so $E$ is the circumcenter of $ACD$. This means $AE=AD$, so $\angle DAE=65$. Finally, $\angle DAC=5$.

Orestis_Lignos wrote: Let $ABC$ be an isosceles triangle, and point $D$ in its interior such that $$D \hat{B} C=30^\circ, D \hat{B}A=50^\circ, D \hat{C}B=55^\circ$$ (a) Prove that $\hat B=\hat C=80^\circ$. (b) Find the measure of the angle $D \hat{A} C$. Let E a point on segment BD, such that $\angle ECB=30$. Point $E$ is on the bicector of angle $A$. It is very easy to see that rays $CD,ED$ are agnle bisectors in triangle $AED$ , so point $D $ in the incenter of triangle $AEC$. The rest is obvious, since $\angle EAC=10^0$.

rama1728 wrote: If I am not wrong, Part a is trivial because \[\angle ABC=\angle DBA+\angle DBC=80\]and since it's isoceles we're done. It is but you have to check the cases $AC=BC$ and $AB=BC$

sttsmet wrote: rama1728 wrote: If I am not wrong, Part a is trivial because \[\angle ABC=\angle DBA+\angle DBC=80\]and since it's isoceles we're done. It is but you have to check the cases $AC=BC$ and $AB=BC$ Actually, there are 3 cases. * $AB$ $=$ $BC$ (sides where B lies are equal) * $AB$ $=$ $AC$ (sides where A lies are equal) * $AC$ $=$ $BC$ (sides where C lies are equal)

sttsmet wrote: rama1728 wrote: If I am not wrong, Part a is trivial because \[\angle ABC=\angle DBA+\angle DBC=80\]and since it's isoceles we're done. It is but you have to check the cases $AC=BC$ and $AB=BC$ But, assuming $AB=BC$, $D$ will be outside triangle, so there is no question!

a) casework