+8 
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?
+128460 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
+9466 
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 
+26367 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
+26367 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?
\(\text{Let $AB=DC=b$ } \\ \text{Let $AD=BC=a$ } \\ \text{Let $DE=BE'=E'C'=x$ } \\ \text{Let $BD=d$ } \)
\(\begin{array}{|rcll|} \hline d^2 &=& a\cdot 2x \\\\ x &=& \dfrac{d^2}{a} \quad & | \quad d^2 = a^2+b^2 \\ &\mathbf{x } &\mathbf{=}& \mathbf{\dfrac{a^2+b^2}{2a}} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline a&=&10 \\ b &=& 5 \\\\ ED=x &=& \dfrac{5^2+10^2}{2\cdot 10} \\ x &=& \dfrac{125}{20} \\ x &=& \dfrac{25}{4} \\ \mathbf{x } &\mathbf{=}& \mathbf{6.25} \\ \hline \end{array}\)
