Triangles (1).

Let the side of the triangles be represented by a, b & c

Therefore length of tangent from vertex A is equal to s-a and similarly s-b and s-c from vertices B & C respectively

Given s-a = 8, s-b = 6, s-c = 3 Adding the three we get

3s-(a+b+c) = 17 therefore s = 17

thus a = 9, b = 11, c = 14

Also length of tangent from vertex A = semi-perimeter of triangle ADE

i.e. 8 = $\frac { \frac { { c }{ .p } }{ a } +\frac { { b }.{ p } }{ a } +{ p } }{ 2 }$

which gives p = $\frac { { 16 }{ a } }{ { a }+{ b }+{ c } }$

therefore q = $\frac { { 12 }{ b } }{ { a }+{ b }+{ c } }$ r = $\frac { { 6 }{ c } }{ { a }+{ b }+{ c } }$

thus $\left| \frac { { p }+{ q }+{ r } }{ { p }-{ q }-{ r } } \right|$ = $\left| \frac { 360 }{ -72 } \right|$ = 5

$c=8+6=14,..............a=6+3=9,............b=8+3=11,\\ s=1/2*(a+b+c)=17 \\ Area~=\sqrt {s*(s-a)*(s-b)*(s-c) }=49.4793.\\ in-radius,r=\dfrac{2*Area} s =2.91043.\\ Altitudes~h_a=\dfrac{2*area} a=\dfrac{ 2*49.4793} 9=10.9954.\\ Altitudes~h_b=\dfrac{2*area} b=\dfrac{ 2*49.4793}{ 11}=8.99624.\\ Altitudes~h_c=\dfrac{2*area} c=\dfrac{ 2*49.4793} {14}=7.06847.\\ \therefore~p=\dfrac{a*(h_a-2r)}{h_a} =\dfrac{9*(10.9954-5.82086)}{10.9954} = 4.23545.\\ \therefore~q=\dfrac{a*(h_b-2r)}{h_b} =\dfrac{9*(8.99624-5.82086)}{8.99624} = 3.88259.\\ \therefore~r=\dfrac{a*(h_c-2r)}{h_c} =\dfrac{9*(7.06847-5.82086)}{7.06847} = 2.47097.\\ So~~~\dfrac{p+q+r}{p-q-r}=\Huge~~~\color{#D61F06}{5}.$