Calculate the segment EC in the rectangle ABCD

Question

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 $

$\dfrac{EC}{\sin(26.5^\circ)} = \dfrac{EL}{\sin(180^\circ - (26.5^\circ + 30^\circ))} \Longrightarrow$

$EC = \displaystyle\frac{\sin(26.5^\circ)(4+2\sqrt{3})}{\sin(123.5^\circ)} \approx{4}$

Answer 1

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.

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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.

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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, $$EC = \frac{\sin(26.5^\circ)(4+2\sqrt3)}{\sin(123.5^\circ)}\approx (4+2\sqrt3)\cdot \frac{2}{2+\sqrt3} = \boxed{4}$$

Answer 2

Hint: For geometric solution consider this construction:

(1):$D_1C_1=tan 60^o=\sqrt 3$

(2): $B_1G=E_1B_1=D_1C_1$

(3):$C$ is reflect of $B_1$ about $G$, so $B_1C=2\sqrt 3\Rightarrow BC=4+2\sqrt 3$

(4):Triangle DEC is isosceles so CD and CE are symmetric about perpendicular bisector CE . $LC=BE$ , this means CM is also a perpendicular bisector o BL and ALCE is an isosceles trapezoid and $ CE||BL$ and $BE_1CE$ is a rhombus.

(5): Points $E$, $B_1$ and $E_1$ are on a circle with radius $4$, hence $CE=BE=4$

As you see the construction shows $\angle ELC=21.09$, you can follow this instruction and make the accurate construction.Also $AB=5.56$.

Answer 3

Here is an exact (as possible) picture in geogebra:

To see that $\Delta CDE$ is equilateral use a reflection denoted by a prime w.r.t. the angle bisector in $D$ of this triangle. Then $ABCD$ is reflected in $KLED$ because it is enough to check that $C,E$ and $A,K$ reflect in each other. From $BE=CK$ we obtain $AE=A'E'=KC=CK=BE$, so $E$ is on the perpendicular bisector of the segment $AB$, thus also of $CD$, giving $EC=ED$. By reflection, $C$ is on the perpendicular bisector of the segment $DE$, so $CE=CD$, making together $\Delta CDE$ equilateral.

This clears the question concerning this triangle and the angles in $C,D,E$.

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It turns out that the exact value of $CE$ is not $4$, but some number "close to four", the value (shown with $15$ decimal digits precision) is the label of $CE$. We can get as many decimals as we want in this century, here are some first few of them:

sage: F.<u> = CyclotomicField(720)

sage: j = u^90     # this is sqrt(-1)
sage: s = (u^53 - u^-53)/2/j    # this is sin(53°/2)
sage: t = (u^247 - u^-247)/2/j    # this is sin(247°/2)
sage: (s/t * (4 + 2*sqrt(3))).n(300).real()

And we obtain: $$ 3.99391107195919450602102403886251977140233475054345819553790569969721788705367742719545068\dots $$ with first geogebra decimals confirmed.

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To have still some "geometric confirmation" that we land "near four", let us make the same construction "using the four", but letting that $\frac 12{53^\circ}$ angle open. And here is the picture, we are grouping a square and an equilateral triangle both with side four together:

and it turns out that $$ \arctan\frac 24 =\arctan\frac12\approx 26.565051177077989351572193720453294671204214299645\dots^\circ $$ is almost $\frac 1253^\circ$. (The points $C,D,E$ from the last picture are approximations of the "real ones".)