In right triangle ABC, with AB = 44 cm and BC = 33 cm, point D lies on side BC so that BD:DC = 2:1. If vertex A is folded onto point D to create quadrilateral BCEF, as shown, what is the area of triangle CDE? Click here for the solution.
Solution 1: Obviously, $\triangle{ABC}$ is a 3-4-5 triangle with $AC=55$.
Because $BD:DC=2:1$, and $BC=33$, we have $BD=22$, $CD=11$.
Let $BF=x$, $AF=DF=AB-BF=44-x$. Because $\triangle{BDF}$ is a right triangle, according to Pythagorean Theorem, we have
$$DF^2=BD^2+BF^2\ \ \ \ i.e.\ \ (44-x)^2=22^2+x^2\tag{1}$$
Solve the above equation, we have $x=\dfrac{33}{2}$. Therefore $BF=\dfrac{33}{2}$, $AF=\dfrac{55}{2}$.
Draw line $AD$ intersecting $EF$ at $G$, and extend $EF$ and $CB$ intersecting at $H$, as shown below. We have $AG\perp EF$ as $\triangle{DEF}$ is a reflection of $\triangle{AEF}$ along line $EF$. $\triangle{AGF}$ is a right triangle.
Therefore, $\triangle{AGF}\sim\triangle{ABD}\sim\triangle{HBF}$. We have $$\dfrac{HB}{BF}=\dfrac{AB}{BD}=\dfrac{44}{22}=2$$ Therefore $HB=2BF=33$, and $CH=HB + BC=66$.
According to Menelaus’s Theorem, we have $$\dfrac{CH}{HB}\cdot\dfrac{BF}{FA}\cdot\dfrac{AE}{EC}=1$$i.e.
$$dfrac{66}{33}\cdot\dfrac{\dfrac{33}{2}}{\dfrac{55}{2}}\cdot\dfrac{AE}{EC}=1$$
Therefore $$\dfrac{AE}{EC}=\dfrac{5}{6}\tag{2}$$
Because $$AE+EC=AC=55\tag{3}$$ Solving (2) and (3), we have: $AE=25$, $EC=30$.
Because $\triangle{DEF}$ is a reflection of $\triangle{AEF}$, $DE=AE=25$. So the lengths of three sides of $\triangle{CDE}$ are $a=11$, $b=25$, and $c=30$. According to Heron’s formula, we have $$s=\dfrac{a+b+c}{2}=33$$ and the area of $\triangle{CDE}$ is
$$S=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{33\times 22\times 8 \times 3}$$ $$=11\sqrt{3\times 2\times 3\times 8}=11\times 3\times 4=\boxed{132}$$
Solution 2: Obviously, $\triangle{ABC}$ is a 3-4-5 triangle with $AC=55$.
Because $BD:DC=2:1$, and $BC=33$, we have $BD=22$, $CD=11$.
Draw line $AD$ intersecting $EF$ at $G$, and draw line $DH$ so that $DH\perp CA$ at $H$, as shown below:
$$AD=\sqrt{AB^2+BD^2}=\sqrt{44^2+22^2}=22\sqrt{5}$$
Because $\triangle{DEF}$ is a reflection of $\triangle{AEF}$, $AG\perp GE$. Therefore, $\triangle{AGE}\sim\triangle{AHD}$ and $AG=\dfrac{1}{2}AD=11\sqrt{5}$.
The area of $\triangle{ACD}$ is $$[\triangle{ACD}]=\dfrac{1}{2}AB\cdot CD=\dfrac{1}{2}\times 44\times 11=242$$
Because $[\triangle{AHD}]=\dfrac{1}{2}AC\cdot DH$, we have $$242=\dfrac{1}{2}\times 55\cdot DH$$
Therefore $DH=\dfrac{44}{5}$.
Because $\triangle{AHD}$ is a right triangle, we have $$AH=\sqrt{AD^2-DH^2}=\sqrt{(22\sqrt{5})^2-(\dfrac{44}{5})^2}=22\sqrt{5-\dfrac{4}{25}}=22\times\dfrac{11}{5}=\dfrac{242}{5}$$
Because $\triangle{AGE}\sim\triangle{AHD}$, we have $$\dfrac{GE}{AG}=\dfrac{DH}{AH}=\dfrac{\dfrac{44}{5}}{\dfrac{242}{5}}=\dfrac{2}{11}$$
Therefore $$EG=\dfrac{2}{11}AG=\dfrac{2}{11}\cdot 11\sqrt{5}=2\sqrt{5}$$
Therefore the area of $\triangle{ADE}$ is $$[\triangle{ADE}]=\dfrac{1}{2}AD\cdot DH=\dfrac{1}{2}\cdot 22\sqrt{5}\cdot 2\sqrt{5}=110$$
The area of $\triangle{CDE}$ is
$$[\triangle{CDE}]=[\triangle{ACD}]-[\triangle{ADE}]=242-110=\boxed{132}$$