Calculate the length of a side of a trapezoid

Question

Let $ABCD$ a trapezoid such that :$(AB)//(CD)$ ; $\widehat{ADC}=120°$ ; $AB=3CD=3DA=12$ Compute $BC$.

My attempt:

I introduced the orthogonal projection of $C$ (named $K$) on the segment $[AB]$ and i managed to show that $\widehat{CAK}=30°$ and $\widehat{ACK}=60°$ . I need now to prove that the triangles $ACK$ and $CKB$ are congruent, then i can compute $BC$ easily

Thank you for your help

Answer 1

$ED=AD\sin 60^\circ=2\sqrt{3}$

$AE=AD\cos 60^\circ=2$

$EF=CD=4$

$BF=AB-AE-EF=12-2-4=6$

$BC=\sqrt{6^2+(2\sqrt{3})^2}=\sqrt{48}=4\sqrt{3}$

Why the orthogonal projection?

Answer 2

Also introduce the orthogonal projection of D (named J) on the segment AB. Then, $\angle ADJ =30$ and $$CK = DJ = AD \cos \angle ADJ =4 \cos 30 = 2\ \sqrt3$$ $$KB = AB -JK - AJ = 12 - 4 - 4\sin 30 = 6$$

Thus,

$$BC = \sqrt{CK^2+KB^2}= 4\sqrt3$$

Answer 3

like $CK$ draw the line $DM$ then $\widehat{ADM}=30^0$ and $\frac{DM}{AD}=cos 30^0$ then $DM=2\sqrt3$ we have $CK=2\sqrt3$. then we have $\frac{AM}{AD}=sin 30^0 \Longrightarrow AM=2$. $MK=CD=4$.then $BK=AB-AM-MK=6$.BY Pythagorean theorem $BC^2=CK^2+BK^2 \Longrightarrow BC=4\sqrt3$.

Answer 4

If you want to show the triangles $\widehat{ACK}$ and $\widehat{CKB}$ are congruent as shown $AK=6$ then $BK=6 \Longrightarrow AK=BK$ $$\widehat{AKC}=\widehat{BKC}=90^0$$ $$CK=CK$$ $$ $$ therefore two triangles are congruent