Right Triangles Galore

Relevant wiki: Similar Triangles - Problem Solving - Medium

From similar triangles, $\frac { 1 }{ AF } +\frac { 1 }{ DC } =\frac { 1 }{ GH }$ . Solving for $DC$ , we get $DC=12$ .

For right triangles, $\sqrt { \left( BF \right) \left( FC \right) } =AF$ and $\sqrt { \left( FC \right) \left( CE \right) } =DC$ . Solving for $BF$ and $CE$ we get $BF=4$ and $CE=16$ .

Area of polygon $AGDEB$ is equal to $\left[ ABC \right] +\left[ DEF \right] -\left[ GFC \right] =\frac { 13\times 6 }{ 2 } +\frac { 25\times 12 }{ 2 } -\frac { 4\times 9 }{ 2 } =171$ .

Since $\triangle$ AFC and $\triangle$ GHC are similar, we get CH = $\frac{2}{3} * 9$ = 6 and FH =3

Since $\triangle$ DCF and $\triangle$ EDF are similar, $\frac{ED}{DF}$ = $\frac{DC}{CF}$ = $\frac{12}{9}$

Since $\triangle$ DCE and $\triangle$ FDE are similar, $\frac{DF}{ED}$ = $\frac{DC}{CE}$ = $\frac{9}{12}$ from above

So we get, CE = 16. Similarly, we can get BF = 4.

(The above can also be obtained by using, $DC^{2} = CE.CF$ and $AF^{2} = BF.CF$ )

Area of polygon AGDEB

= Area $\triangle$ ABC + Area $\triangle$ DEF - Area $\triangle$ GFC

= 39 + 150 - 18 = 171