Black book 6

$\text{All triangle pairs with common vertex have, their area are simply in proportion to their base.} \\ \text{Let area be shown as [XYZ]. Take } \Delta~ [AFE]= k. ~~So~~\Delta ~[ADF]=\frac{.8}{.2}k =4*k \\ \Delta ~[ADE]=5k;\\$ $\Delta ~[DBF]=\frac{.5}{.5} *\Delta ~[ADF]=4*k.\\ Delta ~[EFC]=\frac{.4}{.6} *\Delta ~[AFE]=\frac 2 3*k\\ \dfrac{\Delta ~[ABC]}{\Delta ~[ADE]}=\dfrac {1*1k}{.5*.6k} \\ \therefore~\Delta ~[ABC]=\dfrac {5k} {.3}=\dfrac {50k} 3 \\ \therefore ~ P=\dfrac {4k}{\dfrac {50k} 3}=\dfrac 6 {25}.\\ \therefore ~ Q=\dfrac {\frac {2k}3}{\dfrac {50k} 3}=\dfrac 1 {25}.\\ 2P+10Q=\dfrac {12} {25} + \dfrac {10} {25} =0.88$

$a,b,c\quad are\quad sides\quad of\quad triangle\quad opposite\quad to\quad A,B,C\quad respectively$

AE= $\frac{6b}{10}$ , EB= $\frac{4b}{10}$ , AD=DC= $\frac{c}{2}$ The side lengths as shown in the figure can be found by ratios given.

$\triangle =\frac { 1 }{ 2 } cH\quad (here\quad H\quad is\quad the\quad height\quad of\quad \triangle ABC\quad with\quad base\quad AB)\\ H=\frac { 2\triangle }{ c } \quad \\ sinA=\frac { H }{ b } =\frac { 2\triangle }{ cb } \\ EI=sinA*EA=\frac { 2\triangle }{ cb } *\frac { 6b }{ 10 } =\frac { 6\triangle }{ 5c } \\ In\quad \triangle ADE,\quad FJ\quad ||\quad EI\quad \\ Applying\quad similarity, using DF:FE=8:2\\ FJ=\frac { 8EI }{ 10 } =\frac { 24\triangle }{ 25c } \quad (FJ\quad is\quad height\quad of\quad \triangle DFC)\\ { \triangle }_{ 1 }\quad =\quad \frac { 1 }{ 2 } *\frac { c }{ 2 } *FJ\quad =\quad \frac { 6\triangle }{ 25 } .\\$

In the very similar way, we find the area of triangle(EFB) from FG. After finding FG from DH.

${ \triangle }_{ 2 }\quad =\quad \frac { \triangle }{ 25 } .$

$P=\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 6 }{ 25 } \\ Q=\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 1 }{ 25 }$

2P+10Q=0.88.

We're done!

For generalization the same steps can be followed through variables.