n.e.e.d.h.e.l.p.

Note that, by AAS, triangle EC₁D   is congruent to triangle EAB

C₁ D  = AB  =  DC  = 5

BC₁  = BC  = 10

And BD  = √[C₁D^2 + BC₁^2]  = √ [5^2 + 10^2 ]  = √125

Note that  BE^2   = AE^2 + AB^2   =  AE^2 + 25

And  BE  = √[AE^2 + 25 ]

By the Law of Cosines, we have

AB^2  = AE^2 + BE^2  - 2 (AE * BE) cosAEB

5^2  = AE^2 + (AE^2 + 25) - 2(AE√[AE^2 + 25] )cos (AEB)

25  = AE^2 + AE^2 + 25 - 2 (AE √[AE^2 + 25] )cos(AEB)

[-2AE^2] / [ -2AE√[AE^2 + 25]   = cos(AEB)

AE/ √[AE^2 + 25]  = cos(AEB)    (1)

Note that  AEB  and DEB  are supplemental....so  cos(DEB)  = -cos(AEB)

So....using the Law of Cosines again, we have

BD^2  =  BE^2 + BE^2  - 2(BE^2)(-cos(AEB))

125 = 2BE^2 + 2BE^2(cos(AEB) )

125 = 2[AE^2 + 25]  - 2[AE^2 + 25] os(AEB)

125 = 2AE^2 + 50 - 2[AE^2 + 25] cos(AEB)

[75 - 2AE^2[ / [ 2(AE^2 + 25) ]  = cos(AEB)   (2)

Equate  (1)  and (2)

[75 - 2AE^2 ] / [ 2(AE^2 + 25)] = AE/ √[AE^2 + 25] 

[75 - 2AE^2] /(AE^2 + 25)  = 2AE/ √[AE^2 + 25] 

[75- 2AE^2] / (AE^2 + 25)  = 2AE√[AE^2 + 25] / (AE^2 + 25)

75 - 2AE^2  = 2AE√[AE^2 + 25]     square both sides

5625 - 300AE^2 + 4AE^4   = 4AE^2 [ AE^2 + 25]     simplify

5625 - 300AE^2 + 4AE^4 = 4AE^4 + 100AE^2 

5625 =  400AE^2

5626/400 = AE^2

225/16  = AE^2

15/4  = AE

So

DE  =  AD  - AE

DE  =  10  - 15/4

DE  = 40/4  - 15/4

DE  = 25/4   =  6.25

Let  AE  =  b

Let  DE  =  c

AE + DE  =  AD

b + c  =  10

And by the Pythagorean theorem....

b² + 5²  =  c²

b² + 25  =  c²

Now we can find  c .

b + c  =  10        so        b  =  10 - c     Use this value for  b  in the second equation.

b² + 25  =  c²

(10 - c)² + 25  =  c²

100 - 20c + c² + 25  =  c²

125 - 20c + c²  =  c²

125 - 20c  =  0

125  =  20c

125 / 20  =  c

6.25  =  c       

Rectangle ABCD is folded along BD, and point C lands on C1. BC1 and AD intersect at point E.

AB=5, AD=10.

What is the length of DE?

$\text{Let $AB=DC=b$ } \\ \text{Let $AD=BC=a$ } \\ \text{Let $DE=x$ } \\ \text{Let $CC'=p$ } \\ \text{Let $EE'=y$ } \\ \text{Let $BD=d$ }$

$\begin{array}{|lrcll|} \hline (1) & \mathbf{d} &\mathbf{=}& \mathbf{\sqrt{a^2+b^2}} \\ \hline & b^2 &=& p\cdot a \\ (2) & p &=& \dfrac{b^2}{a} \\ \hline & y^2 &=& p(p-a) \\ & &=& \dfrac{b^2}{a}\left(\dfrac{b^2}{a}-a \right) \\ & &=& \dfrac{b^2}{a}\left(\dfrac{a^2+b^2}{a} \right) \\ & &=& \dfrac{b^2}{a^2}\left(a^2+b^2\right) \\ (3) &\mathbf{y } &\mathbf{=}& \mathbf{\dfrac{b}{a}\sqrt{a^2+b^2}} \\ \hline & x^2 &=& \left(\dfrac{d}{2}\right)^2 + \left(\dfrac{y}{2}\right)^2 \\ & &=& \left(\dfrac{\sqrt{a^2+b^2}}{2}\right)^2 + \left(\dfrac{\dfrac{b}{a}\sqrt{a^2+b^2}}{2}\right)^2 \\ & &=& \dfrac{a^2+b^2}{4} + \dfrac{b^2}{a^2}\cdot \dfrac{(a^2+b^2)}{4} \\ & &=& \dfrac{a^2+b^2}{4} \left( 1+ \dfrac{b^2}{a^2}\right) \\ & &=& \dfrac{(a^2+b^2)^2}{4a^2} \\ &\mathbf{x } &\mathbf{=}& \mathbf{\dfrac{(a^2+b^2)}{2a}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline a&=&10 \\ b &=& 5 \\\\ DE=x &=& \dfrac{5^2+10^2}{2\cdot 10} \\ x &=& \dfrac{125}{20} \\ x &=& \dfrac{25}{4} \\ \mathbf{x } &\mathbf{=}& \mathbf{6.25} \\ \hline \end{array}$

The length of DE is 6.25