For reference: On a rectangular trapezoid $ABCD$ right at A and B. Mark a point $L$ on $AB$ such that: $CL=AL=AD$ and $AC=CD$, calculate the measure of angle $ACD$.(Answer:$30^o$)
My progresss: Follow the figure with the distribution of angles that I saw
$\triangle ALC: \frac{sen180-2\theta}{AC}=\frac{sen \theta}{AL=AD}\\ \triangle BLC: \frac{sen90}{LC=AD}=\frac{1}{AD}=\frac{sen 2\theta}{BC}\\ \triangle ACD: \frac{sen 90-\theta}{AC}=\frac{sen2\theta}{AD} $
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