In a rectangle $ABCD$, a point $E$ is marked inside it and a rectangle $DELK$ is constructed congruent to the previous one, $m(\angle CED)=m(\angle ECD),\ BE=CK,\ EL=4+2\sqrt3$ and $m(\angle ELC)=\dfrac{53^\circ }{2}$. Calculate $EC$. (Answer:$ EC = 4$)
Would anyone have a solution using geometry? I made the drawing and I find a solution by use of trigonometry
$EC =ED = DC$ ???? How to demonstrate?
$\therefore \angle DEC = 60^\circ \implies \angle LEC = 30^\circ $
$\dfrac{EC}{\sin(26.5^\circ)} = \dfrac{EL}{\sin(180^\circ – (26.5^\circ + 30^\circ))} \Longrightarrow$
$EC = \displaystyle\frac{\sin(26.5^\circ)(4+2\sqrt{3})}{\sin(123.5^\circ)} \approx{4}$
Best Answer
Here is a slick proof of $EC=ED=DC$: (other than that, I don't see any way to avoid trig, because we have a not-so-easy angle measure)
Reflect the rectangle $ABCD$ about $CD$ to get rectangle $A'B'CD$. Let the corresponding reflection of $E$ about $CD$ be $E'$.
Now, rotate the new rectangle $A'B'CD$ about $D$ with an angle of $\alpha$ such that $C$ lands on $E$ (this is possible since we are given $DC = DE$ - in terms of angles). Then $A'$ lands on $K$ and $B'$ lands on $L$. Note that $\angle EDC = \alpha$.
Since $E'$ is the reflection of $E$ about $CD$, the angles they both make at $D$ w.r.t to $CD$ must be equal, i.e, $\angle CDE' = \angle EDC = \alpha$.
This shows that our rotation guarantees that $E'$ lands on $C$.
Next, $BE = CK$ is given, and $CK = E'A'$ by our rotation, so that $BE = E'A' = EA$. This means the perpendicular bisector of $AB$ (which is also perpendicular bisector of $CD$) passes through $E$, so that $CE = DE$ as desired.
Now we can use the sine rule like you did to get: $$EC = \frac{\sin(26.5^\circ)(4+2\sqrt3)}{\sin(123.5^\circ)}\approx \boxed{4}$$Although we can use the well-known $\sin(53^\circ) = 4/5$ to approximate this, I am sure this is very messy without a calculator.
Edit: Actually, we can do the calculation without a calculator:
First, note that $\sin(30^\circ+\theta)/\sin(\theta) = \cot(\theta)/2+\sqrt3/2$. Then $\sin(123.5^\circ)/\sin(26.5^\circ) = \sin(30^\circ+26.5^\circ)/\sin(26.5^\circ) = \cot(26.5^\circ)/2+\sqrt3/2$.
We can find $\cot(26.5^\circ)$ using $$3/4 \approx \cot(26.5^\circ \cdot 2) = \frac{\cot^2(26.5^\circ)-1}{2\cot(26.5^\circ)} = \frac{m^2-1}{2m}$$which, upon solving gives $m \approx 2$ or $-0.5$. The negative value is discarded, hence $\sin(123.5^\circ)/\sin(26.5^\circ) \approx 1+\sqrt3/2 = \frac{2+\sqrt3}{2}$.
Finally, $$EC = \frac{\sin(26.5^\circ)(4+2\sqrt3)}{\sin(123.5^\circ)}\approx (4+2\sqrt3)\cdot \frac{2}{2+\sqrt3} = \boxed{4}$$