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"In the isosceles trapezoid $ABCD$ with bases $AD$ and $BC$ (being $AD$ the larger base), $\angle$ABC = $2$$\angle$CDA. If $AB$ = $BC$ = $b$, the perimeter of the trapezoid is?"

I thought that this problem was very easy, but i can't figure it out. The main problem is the length of the base $AD$.

I tried with similarity of some triangles that are formed with the heights and the diagonals of the trapezoid, but nothing.

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The perimeter is $5b$.

Note: $$mABC+mDAB=180 \Rightarrow mDAB=60.$$ Drop heights $BE$ and $CF$, then: $$\cos{(mBAE)}=\frac{AE}{AB} \Rightarrow AE=b\cos{60}=\frac{b}{2}.$$ Hence: $$P=AB+BC+CD+(AE+DF)+EF=5b.$$

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