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**red_dog** $\displaystyle \widehat{BDC}=\widehat{BCD}=3x\Rightarrow\triangle BDC$ isosceles.

Let $\displaystyle AB=DC=a, \ BE\perp DC, E\in (DC)$ and $\displaystyle F$ the midpoint of $\displaystyle AB$.

Then $\displaystyle FE=DE=EC=\frac{a}{2}\Rightarrow \widehat{DFC}=90$

$\displaystyle \widehat{FEC}=2x, \ \widehat{EFC}=\widehat{FCE}=x$

We have $\displaystyle \widehat{FBD}=\widehat{FCD}=x\Rightarrow FBCD$ is cyclic $\displaystyle \Rightarrow \widehat{DBC}=\widehat{DFC}=90$.

Then $\displaystyle 3x=45\Rightarrow x=15$

(What program do you use for drawing figures?)