Another one of my Triangle Questions
In \(\Delta ABC\), line segments are drawn parallel to each of the sides dividing the triangle into six regions.
The areas of three regions are shown in the figure. What is the area of \(\Delta ABC\)?
\(\begin{array}{l} \text{Let Area $A =[ABC] $} \\\\ \text{Let Area $A_4=[E'EP] =a^2 = 4, ~a=2 $}\\ \text{Let Area $A_5=[D'DP] =b^2 = 9, ~b=3 $}\\ \text{Let Area $A_6=[F'FP] =c^2 = 16 ~c=4$} \\\\ \text{Let Area $A_1=[E'AD] $}\\ \text{Let Area $A_2=[D'BF] $}\\ \text{Let Area $A_3=[F'CE] $} \end{array}\)
\(\text{The triangles are similar $\\( \triangle ABC \sim \triangle E'EP \sim \triangle D'DP \sim \triangle F'FP \sim \triangle E'AD \sim \triangle D'BF \sim \triangle F'CE )$.} \)
The key theorem we apply here is that the ratio of the areas of 2 similar triangles is
the ratio of a pair of corresponding sides squared.
\(\begin{array}{|rcll|} \hline \dfrac{A_4}{A} &=& \left(\dfrac{E'P}{CB}\right)^2 \\ \dfrac{A_5}{A} &=& \left(\dfrac{PD}{CB}\right)^2 \\ \hline \dfrac{A_4}{A}\cdot \dfrac{A}{A_5} &=& \dfrac{E'P^2}{CB^2} \cdot \dfrac{CB^2}{PD^2} \\ \dfrac{A_4}{A_5} &=& \dfrac{E'P^2}{PD^2} \\ \dfrac{a^2}{b^2} &=& \dfrac{E'P^2}{PD^2} \\ \dfrac{a}{b} &=& \dfrac{E'P}{PD} \\ \boxed{\dfrac{E'P}{PD}=\dfrac{a}{b} } \\ \hline \end{array} \)
\(\begin{array}{|rclcrl|} \hline \dfrac{E'P}{PD} &=& \dfrac{a}{b} && E'P &=& \dfrac{a}{a+b}E'D \\ && && E'D &=& \dfrac{a+b}{a} E'P \\ \dfrac{E'D}{PD} &=& \dfrac{a+b}{a} \cdot \dfrac{E'P}{PD} \\ \dfrac{E'D}{PD} &=& \dfrac{a+b}{a} \cdot \dfrac{a}{b} \\ \boxed{\dfrac{E'D}{PD} = \dfrac{a+b}{b} } \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline \dfrac{A_1}{A_5} &=& \left( \dfrac{E'D}{PD} \right)^2 \\ A_1 &=& A_5\left( \dfrac{E'D}{PD} \right)^2 \\ A_1 &=& b^2\left( \dfrac{a+b}{b} \right)^2 \\ \mathbf{A_1} & \mathbf{=} & \mathbf{(a+b)^2} \quad & | \quad A_1 = (2+3)^2=5^2=25 \\ \hline \end{array} \)
analogous
\(\begin{array}{|rclcrl|} \hline \dfrac{D'P}{PF} &=& \dfrac{b}{c} && D'P &=& \dfrac{b}{b+c}D'F \\ && && D'F &=& \dfrac{b+c}{b}D'P \\ \dfrac{D'F}{PF} &=& \dfrac{b+c}{b} \cdot \dfrac{D'P}{PF} \\ \dfrac{D'F}{PF} &=& \dfrac{b+c}{b} \cdot \dfrac{b}{c} \\ \boxed{\dfrac{D'F}{PF} = \dfrac{b+c}{c} } \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \dfrac{A_2}{A_6} &=& \left( \dfrac{D'F}{PF} \right)^2 \\ A_2 &=& A_6\left( \dfrac{D'F}{PF} \right)^2 \\ A_2 &=& c^2\left( \dfrac{b+c}{c} \right)^2 \\ \mathbf{A_2} & \mathbf{=} & \mathbf{(b+c)^2} \quad & | \quad A_2 = (3+4)^2=7^2=49 \\ \hline \end{array}\)
analogous
\(\begin{array}{|rclcrl|} \hline \dfrac{F'P}{PE} &=& \dfrac{c}{a} && F'P &=& \dfrac{c}{a+c}F'E \\ && && F'E &=& \dfrac{a+c}{c}F'P \\ \dfrac{F'E}{PE} &=& \dfrac{a+c}{c} \cdot \dfrac{F'P}{PE} \\ \dfrac{F'E}{PE} &=& \dfrac{a+c}{c} \cdot \dfrac{c}{a} \\ \boxed{\dfrac{F'E}{PE} = \dfrac{a+c}{a} } \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \dfrac{A_3}{A_4} &=& \left( \dfrac{F'E}{PE} \right)^2 \\ A_3 &=& A_4\left( \dfrac{F'E}{PE} \right)^2 \\ A_3 &=& a^2\left( \dfrac{a+c}{a} \right)^2 \\ \mathbf{A_3} & \mathbf{=} & \mathbf{(a+c)^2} \quad & | \quad A_3 = (2+4)^2=6^2=36 \\ \hline \end{array} \)
\(\mathbf{A=\ ?}\)
\(\begin{array}{|rcll|} \hline A_1+A_2+A_3 &=& \mathbf{A}+ A_4+A_5 + A_6 \\ 25+49+36 &=& A+ 4+9+16 \\ 110 &=& A + 29 \\ A &=& 110 - 29 \\ \mathbf{A} & \mathbf{=} & \mathbf{81} \\ \hline \end{array}\)
+583 
