Cyclic quadrilateral $ABCD$ has lengths $BC=CD=3$, $AB=5$, $AD=8$. What is the length of the shorter diagonal of $ABCD$? Click here for the solution.
Solution: Let $AC$ and $BD$ intersect at $E$, $x=BD$, $y=AE$, $z=CE$, $AC=y+z$.
Because $BC=CD$, $AC$ is angle bisector of $\angle{BAD}$. By Angle Bisector Theorem, we have $\dfrac{BE}{DE}=\dfrac{AB}{AD}=\dfrac{5}{8}$. Because $BE+DE=BD=x$, we have $BE=\dfrac{5}{13}x$, $DE=\dfrac{8}{13}x$.
By Ptolemy’s Theorem, we have $BD\cdot AC=AB\cdot CD+AD\cdot BC$. Hence, $$x(y+z)=5\cdot 3+3\cdot 8=39\tag{1}$$
Because $BC=CD$, we have $\angle{BAC}=\angle{DAC}=\angle{BDC}=\angle{CAD}$.
Because $\angle{BCE}=\angle{ADE}$, $\triangle{BCE}\sim\triangle{ADE}$. Therefore $$\dfrac{CE}{DE}=\dfrac{BE}{AE}=\dfrac{BC}{AD}\ \ \ \ i.e.\ \ \ \ \dfrac{z}{\dfrac{8}{13}}=\dfrac{\dfrac{5}{13}x}{y}=\dfrac{3}{8}$$ We have $$y=\dfrac{40}{39}x\tag{2}$$ $$z=\dfrac{3}{13}x\tag{3}$$
Solving equation (1), (2) and (3), and ignoring negative solutions, we have $$x=\dfrac{39}{7}\ \ \ \ \ \ \ y=\dfrac{40}{7}\ \ \ \ \ \ z=\dfrac{9}{7}$$
Therefore, $BD=x=\dfrac{39}{7}$, $AC=y+z=7$. Therefore, the length of the shorter diagonal is $\boxed{\dfrac{39}{7}}$.
Note: since $\triangle{BCE}\sim\triangle{ACB}$, we have $\dfrac{BC}{CE}=\dfrac{AC}{BC}$, i.e. $\dfrac{3}{z}=\dfrac{y+z}{3}$. We have $$z(y+z)=9\tag{4}$$ We obtain the same result by solving equation (2), (3) and (4).