As above, we begin be finding all other relevant side lengths: [asy]
unitsize(1inch);
pair C, D, E, F, A, B;
C = (0,0);
D = (1.73,0);
F = (1.73/2,1.73);
A = rotate(60)*(1.73/3,0);
B = rotate(60)*(1.73/3,1.73/3);
E = rotate(60)*(1.73/3,2*1.73/3);
draw(A--F--D--B--C);
label("$A$",A,NW);
label("$B$",B,N);
label("$C$",C,W);
label("$D$",D,E);
label("$E$",E,NE);
label("$F$",F,S);
label("$3$",(D+C)/2,S);
label("$3$",(B+D)/2,NE);
label("$5$",(E+D)/2,N);
[/asy] Since $\triangle CDE$ has all sides specified, it is an isosceles triangle.
Since $CD=CE$, $\angle CDE = \angle CED$.
Since the angle sum of $\triangle CDE$ must be $180^\circ$, $\angle D = 90^\circ$. Thus we conclude that $\triangle CED$ is a 45-45-90 triangle, so $DE = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}$.
We know that $\triangle CBF$ is a 30-30-120 triangle, and since we labeled a few side lengths, we should be able to find some additional ones fairly easily:
We know that $BF=\frac{AF}{2}$ and $AB=BF$, therefore, $AB=\frac{AF}{2}$.
We know that $FD=2 \cdot DE = \frac{10\sqrt{2}}{2}$, and $AD = DE -AB = \frac{5\sqrt{2}}{2} - \frac{AF}{2}$.
Applying the Pythagorean Theorem to find $AF$:
$AF^2 = AD^2 + DF^2 = (\frac{5\sqrt{2}}{2} - \frac{AF}{2})^2 + (\frac{10\sqrt{2}}{2})^2$.
$AF^2 = (\frac{5\sqrt{2}}{2} - \frac{AF}{2})^2 + (\frac{10\sqrt{2}}{2})^2 = \left( \frac{5\sqrt{2}}{2} - \frac{AF}{2} \right) \left( \frac{5\sqrt{2}}{2} - \frac{AF}{2} \right) + \left( \frac{10\sqrt{2}}{2} \right) \left( \frac{10\sqrt{2}}{2} \right)$. $AF^2 = \left( \frac{15}{2} - \frac{AF}{2} \right) \left( \frac{5}{2} - \frac{AF}{2} \right) + \frac{100}{2}$.
$AF^2 = \left( 6 - AF \right) \left( 2 - AF \right) + 50$.
$AF^2 = 12 - 8AF + AF^2 + 4AF - AF^2 + 50$.
$0 = 22 - 4AF$.
$4AF = 22$, resulting in $AF = \frac{11}{2}$.
Having found $AF$, it is now a matter of using the Pythagorean Theorem one more time to find $AE$:
$AF^2 = AE^2 + EF^2$.
$\left( \frac{11}{2} \right)^2 = AE^2 + \left( \frac{5\sqrt{2}}{2} \right)^2$.
$\frac{121}{4} - \frac{50}{2} = AE^2$.
$AE^2 = \frac{121 - 100}{4}$.
$AE^2 = \frac{21}{4}$.
$AE = \boxed{\frac{\sqrt{21}}{2}}$
The vertices of triangle $ABC$ lie on the sides of equilateral triangle $DEF$, as shown. If $CD = 5$, $CE = BD = 3$, and $\angle ACB = 90^\circ$, then find $AE$.
1 answer