Twisted Restraints

Relevant wiki: Length and Area Problem Solving

$\text{Area} \ ABCD = 672$

$\text{Area} \ ACB = \frac{672}{2} = 336$

$AE = \frac{336 \times 2 }{42} = 16$

Now we can add one more diagonal

$AG = GD = 21$

Step into Pythagorean theorem

$AE^2 + EG^2 = AG^2$

$16^2 + EG^2 = 21^2$

$EG = GF = \sqrt{185}$

Looking at the figure above, we can conclude that

$CE = \frac{42 - 2\sqrt{185}}{2}$

$CE = \frac{2(21-\sqrt{185})}{2}$

$CE = 21 - \sqrt{185}$

$CE \approx 7.4$

Here's a less clever method based on similar triangles. First, note that we can express $A_{\triangle ACB}$ in two ways: Thanks to diagonal symmetry, $A_{\triangle ACB} = \frac{672}{2} = 336$ . But since $AE\perp CB$ , we also know that $A_{\triangle ACB} = \frac{1}{2}\cdot 42\cdot AE = 336$ . Thus, $AE = 16$ .

Now, since $\angle CBD \cong \angle ACB$ , we know $\angle CAE \cong \angle CBA$ . Thus, $\triangle ACE\sim\triangle BAE$ , and it follows that $\frac{CE}{AE} = \frac{AE}{CB - CE}$ $\frac{CE}{16} = \frac{16}{42 - CE}$ $CE^2 - 42CE + 256 = 0.$ From the quadratic formula, we find $CE \approx 7.4$ or $CE \approx 34.6$ . But since $42 = CE + EF + FB = 2CE + EF$ , in order for $EF$ to be non-negative, we must reject the larger solution and conclude $CE \approx 7.4.$

$Let\ AC=x \ <\ AB=y.\\ x^{2}+y^{2}=42^2,\ \ \ \ \ \ xy=672.\ \ \ \\ Also\ BC*AE=672,\ \ \ \ \ \implies\ AE=\dfrac{672}{42}=16.\\ \therefore\ (x+y)^2=3108\ \ \ \ \ \ \ (x-y)^2=420.\\ \implies\ x=\sqrt{777}-\sqrt{105},\\ CE^2=AC^2- AE^2=x^2-16^2=54.738.\\ So\ CE=7.398$