MathFun4Kids

Solution 1 Draw line $\overline{BE}$ intersecting with line $\overline{AC}$ at $F$. Obviously, quadrilateral $CDEF$ is a rhombus, and both $\triangle{BAF}$ and $\triangle{BEA}$ are similar isosceles triangles.

Let $\overline{AB}=\overline{AE}=\overline{EF}=x$, $\overline{AF}=\overline{BF}=y$, $$\because \triangle{BAF}\sim\triangle{BEA}$$ $$\therefore \dfrac{\overline{BA}}{\overline{BF}}=\dfrac{\overline{BE}}{\overline{BA}}\tag{1}$$ $$\dfrac{x}{y}=\dfrac{x+y}{x}=\dfrac{1}{1+\dfrac{x}{y}}$$ Solve the above equation for $\dfrac{x}{y}$ and ignore the negative value, we have $$\dfrac{x}{y}=\dfrac{\sqrt{5}+1}{2}\tag{2}$$

Let $G$ be the midpoint of $\overline{AF}$. Since $M$ is also the midpoint of $\overline{AE}$, we have $\overline{MF} \parallel \overline{BE}$, therefore $$\triangle{AMZ}\sim\triangle{AEF}$$ Because $\overline{AE}=\overline{EF}=x$, therefore $$ \overline{MA}=\overline{MG}=\dfrac{\overline{AE}}{2}=\dfrac{x}{2}$$ Additionally $$\because \triangle{MGZ}\sim\triangle{BFZ}$$ $$\therefore \dfrac{\overline{ZG}}{\overline{ZF}}=\dfrac{\overline{MA}}{\overline{BF}}=\dfrac{\dfrac{x}{2}}{y}=\dfrac{1}{2}\cdot\dfrac{x}{y}$$ Therefore $$\overline{ZG}+\overline{ZF}=\dfrac{1}{2}\cdot\dfrac{x}{y}\cdot\overline{ZF}+\overline{ZF}=(\dfrac{1}{2}\cdot\dfrac{x}{y}+1)\cdot\overline{ZF}$$ $$\because \overline{ZG}+\overline{ZF}=\overline{GF}=\dfrac{y}{2}$$ $$\therefore (\dfrac{1}{2}\cdot\dfrac{x}{y}+1)\cdot\overline{ZF}=\dfrac{y}{2}$$ $$\because \overline{ZF}=\overline{AF}-\overline{AZ}=y-3$$ $$\therefore (\dfrac{1}{2}\cdot\dfrac{x}{y}+1)\cdot(y-3)=\dfrac{y}{2}\tag{5}$$

Solve equation (2) and (5), we have $$y=\dfrac{15-3\sqrt{5}}{2}$$

And finally we have the answer: $$\overline{AB}=x=\dfrac{x}{y}\cdot{y}=\dfrac{\sqrt{5}+1}{2}\cdot\dfrac{15-3\sqrt{5}}{2}=3\sqrt{5}$$

Solution 2 – Use Mass Points Method Follow the same process as Solution 1 to obtain the result shown in (2).

Assign $A_{mass}=1$, because $M$ is the midpoint of side $\overline{AE}$, we have $$E_{mass}=1, \ \ \ M_{m a s s}=A_{mass}+E_{mass}=2$$ $$\because \dfrac{\overline{EF}}{\overline{BF}}=\dfrac{x}{y}$$ $$\therefore B_{mass}=\dfrac{x}{y}$$ We have $$F_{mass}=E_{mass}+B_{mass}=1+\dfrac{x}{y}\tag{6}$$ and balanced mass at $Z$ as $$Z_{mass}=M_{mass}+B_{mass}=A_{mass}+F_{mass}=2+\dfrac{x}{y}$$ Therefore $$A_{mass}\cdot\overline{AZ}=F_{mass}\cdot\overline{FZ}=F_{mass}\cdot(\overline{AF}-\overline{AZ})$$ i.e. $$1\cdot 3=(1+\dfrac{x}{y})(y-3)\tag{7}$$

Solve equation (2) and (7), we have $$y=\dfrac{15-3\sqrt{5}}{2}$$

And finally we have the answer: $$\overline{AB}=x=\dfrac{x}{y}\cdot{y}=\dfrac{\sqrt{5}+1}{2}\cdot\dfrac{15-3\sqrt{5}}{2}=3\sqrt{5}$$

Solution 3 – Use Menelaus‘ theorem Follow the same process as Solution 1 to obtain the result shown in (2).

According to Menelaus’ theorem, we have the following:

$$\dfrac{\overline{EB}}{\overline{BF}}\cdot\dfrac{\overline{FZ}}{\overline{AZ}}\cdot\dfrac{\overline{AM}}{\overline{EM}}=1$$

Since $\overline{FZ}=\overline{AF}-\overline{AZ}$, we have

$$\dfrac{x+y}{y}\cdot\dfrac{y-3}{3}\cdot\dfrac{\dfrac{x}{2}}{\dfrac{x}{2}}=1$$

i.e.

$$ (1+\dfrac{x}{y})\cdot(y-3)=3\tag{8}$$

Solve equation (2) and (8), we have $$y=\dfrac{15-3\sqrt{5}}{2}$$

And finally we have the answer: $$\overline{AB}=x=\dfrac{x}{y}\cdot{y}=\dfrac{\sqrt{5}+1}{2}\cdot\dfrac{15-3\sqrt{5}}{2}=3\sqrt{5}$$

Hint Draw two red lines as shown