could any smart person help solve this question?
thank you so much.....
Find x where AB=BC
This is not a hard question, it is an impossible one. Not enough information here. Now I could be missing something, but I don't see how you can lock in an angle for x. If m<DAC = 0, then x=90. But if m<DAC = 30, then x = 60. And if m<DAC = 45, then x = 45. This is true regardless of what is going on with triangleABC. We need some other constraint on the problem.
It is a messy problem, but not impossible.
To simplify notation, let $\displaystyle m\left( {\angle DAC} \right) = y\;\& \,m\left( {\angle BAC} \right) = m\left( {\angle BCA} \right) = z$
The sum of measures of angles of a quadrilateral is [$\displaystyle 360^o$.
So $\displaystyle x+y+2z+205=360$. But $\displaystyle x+y=90$, so $\displaystyle 2z+295=360$.
Now you can finish.