Using the Properties of Similar triangles and angle properties to find the length AC

Question

Find the length of AC

What i tried

Since the diagram is a trapezoid, angle $DAB$ is $180-140=40$ degrees. Then using the properties of simillar triangles, angle $DAC$ is equal to angle $CAB$,thus angle $DAC=40/2=20$ degrees.

Then using the sine rule we can find length $AC$ using the formula,

$$\frac{AC}{sin140}=\frac{3}{sin20}$$

Am i correct. Could anyone explain. Thanks

Answer 1

We see that $\angle ACB=\angle CDA$ and $\angle BAC=\angle ACD$, therefore $\triangle ABC\sim \triangle CAD$. Thus we have $$\frac{AC}{AB}=\frac{DC}{AC}\Rightarrow AC^2=12\cdot 3\Rightarrow AC=6$$