Don't get trapped

Construct a Line Perpendicular to CD through A and let the point where that line and CD intersect be E. Similarly do the same through B and let the point be F. Let $AB=x, DE=y.$ By the fact that this trapezoid is isoceles DE is equal to FC. Thus we have $AB=x, DE=y, EC=x+y, DC=x+2y$ It is stated that CD is equal to AC and since AE is perpendicular to CD we can set up the pythagorean theorem. $AE^{2}+EC^{2}=AC^{2}$ $x^2+(x+y)^2=(x+2y)^2$ $x^2+x^2+2xy+y^2=x^2+4xy+4y^2$ Solving this equation: $x^2-2xy-3y^2=0$ $(x-3y)(x+y)=0$ $x=3y, x=-y$ Negative values are extraneous so we have x=3y, AB=3y, and CD=5y. Thus the ratio of 3:5. And the soultion of 243.

$\text{Let n=AB= AE and 2m+n=CD=AC. See figure.}\\ \text{ By Pythagoras }AD=\sqrt{m^2+n^2}, half ~the ~isosceles~ base =\frac{ \sqrt{m^2+n^2} }{2}. \\In~ ~rt \angle d~ \triangle s ~ADE~ and~ CDF, ~~~~~~~ \dfrac{m}{ \sqrt{m^2+n^2} }=Cos \alpha = \dfrac{ \frac{ \sqrt{m^2+n^2} } {2} } {2m+n}\\$ $~~\implies 4m^2+2mn=m^2+n^2\\Solving ~quadratic~3m^2+2mn-n^2=0,~~m=\dfrac n 3 .~~ \therefore n=3m. \\ \dfrac{AB}{CD}= \dfrac{n}{2m+n}=\dfrac{3m}{2m+3m} =\dfrac 3 5 =\dfrac a b \implies a^b=3^5 = \color{#D61F06}{ \large \boxed{243} }$

Awesome Problem , I found very interesting Properties in this question ..

Okay , First Take advantage of fact given in question , Let $AB=a$ & $CD=b$ ,

Now Key is to Switch to co-ordinate geometry .. Let us choose sutaible co-ordinates for above stated conditions , $D(0,0)\& C(b,0)\& \\ A(c,a)\& B(c+a,a)$ .

Note : I will repetedly use distance formula and square it in both sides

$\displaystyle{(i)\quad \because AC=CD\\ { (c-b) }^{ 2 }+{ a }^{ 2 }={ b }^{ 2 }\\ \boxed { { c }^{ 2 }+{ a }^{ 2 }=2bc } \\ (ii)\quad \because AD=BC(\because Isosclace\quad trap.)\\ { c }^{ 2 }+{ a }^{ 2 }={ (c+a-b) }^{ 2 }+{ (a) }^{ 2 }\\ 0=2ac+({ a }^{ 2 }+{ b }^{ 2 }-2ab)-2bc\\ { (b-a) }^{ 2 }=2c(b-a)\quad \quad (\because b\neq a)\\ \boxed { b+a=2c } }$

Now isn't it is beuatiful that $a,c,b$ are in AP (Airthmetic Progression) ?

Now from above two conditions (on which i had putted Box) Put value of C from 2nd Box in to 1st box , and then we devide by $b^2$ to make quadaratic in $a/b$ ..

$\displaystyle{\because { c }^{ 2 }+{ a }^{ 2 }=2bc\\ \cfrac { { (a+b) }^{ 2 } }{ 4 } +{ a }^{ 2 }=(b+a)b\\ 5{ a }^{ 2 }+2ab-3{ b }^{ 2 }=0\\ \cfrac { a }{ b } =\lambda \quad (Let)\\ 5{ \lambda }^{ 2 }+2\lambda -3=0\\ \lambda =\cfrac { 3 }{ 5 } ,-1(\times )\\ \boxed { \cfrac { a }{ b } =\cfrac { 3 }{ 5 } } }$

Without loss of generality let $|CD| = 1.$ Now let $P$ be the point on $CD$ that is the base of the altitude to vertex $A.$ Letting $|AB| = r,$ we have that

$|DP| = \frac{1}{2}(1 - r) \Longrightarrow |PC| = |DC| - |DP| = 1 - \frac{1}{2}(1 - r) = \frac{1}{2}(1 + r).$

Applying Pythagoras to $\Delta APC,$ we find that

$|AC|^{2} = |AP|^{2} + |PC|^{2} \Longrightarrow 1 = r^{2} + \frac{1}{4}(1 + r)^{2}$

$\Longrightarrow 4 = 4r^{2} + 1 + 2r + r^{2} \Longrightarrow 5r^{2} + 2r - 3 = 0 \Longrightarrow (5r - 3)(r + 1) = 0.$

Since $r \gt 0$ we find that $r = |AB| = \frac{3}{5}.$ But with $|CD| = 1,$ $r$ is the ratio we are wanting to find, and thus $\large a^{b} = 3^{5} = \boxed{243}.$

An easy solution.

In right-angled $\triangle EAC$ ,

$a^2 + (\frac{b+a}{2})^2 = b^2$

Divide both sides by $b^2$

$\Rightarrow (\frac{a}{b})^2 + \frac{1}{4}(1+\frac{a}{b})^2 = 1$

let $\frac{a}{b} = \alpha$

$\Rightarrow \alpha^2 + \frac{1}{4}(1+\alpha)^2 = 1$

$\Rightarrow \frac{1}{4}(1+\alpha)^2 = 1 - \alpha^2$

Since $\alpha \neq -1$

$\Rightarrow \frac{1}{4}(1+\alpha) = 1 - \alpha$

$\Rightarrow 1+\alpha = 4 - 4\alpha$

$\Rightarrow \alpha = \frac{3}{5}$

$\therefore a = 3$ , $b =5$ , and $a^b = 243$

From $A$ drop a perpendicular onto $DC$ and let it intersect $DC$ at $L$ . Drop another perpendicular from $B$ onto $DC$ and let it intersect $DC$ at $M$ .

So we have $LM = AB$

Now, $CD = DL + LM + MC = 2DL + AB$ implying $DL = \frac{CD-AB}{2}$

So we now have, $CL = CD - DL = \frac{AB+CD}{2}$

In the right $\Delta ALC$ ,

$AC^2=AL^2+CL^2$

Putting $AC=CD$ and $AL=AB$ in the above equation

$CD^2 =AB^2+(\frac{AB+CD}{2})^2$

Divding both sides by $CD^2$ , taking $y = \frac{AB}{CD}$ and simplifying we get

$5y^2+2y-3=0$

$(5y-3)(y+1)=0$

$y=\frac{3}{5}$ (since $y>0$ )

So we have $\frac{a}{b} = \frac{3}{5}$ so $a^b = 3^5 = \boxed{243}$

Sorry, it should be trng ACE and not ADE

Let $|AB| = a$ and $|CD|=b$ , therefore $\dfrac {|AB|}{|CD|} = \dfrac {a}{b}$ , which is what we want to find. Let the foot of altitude from $A$ to $DC$ be $E$ .

We note that $|AB|=|AE|=a$ , $\space |CD|=|AC| = b$ and $|CE| = b - \dfrac {b-a}{2} = \dfrac {b+a}{2}$ .

We also note that

$|CE|^2+|AE|^2 = |AC|^2\quad \Rightarrow \left( \dfrac {b+a}{2} \right)^2 + a^2 = b^2 \\ \Rightarrow b^2 + 2ab + a^2 + 4a^2 = 4b^2\quad \Rightarrow 3b^2 - 2ab - 5a^2 = 0 \\ \Rightarrow (3b-5a)(b+a) = 0 \quad \Rightarrow 3b=5a \quad \Rightarrow \dfrac {a}{b} = \dfrac{3}{5} \\ \Rightarrow a^b = 3^5 = \boxed{243}$