User:Prophet of Why

For various proofs in mathematics

Concurency
${\displaystyle {\begin{array}{clcl}{\overline {AB}}&y=0&{\overline {AD}}&y={a \over c+b}x\\{\overline {BC}}&y={a \over c+b}x-{ab \over c-b}&{\overline {BE}}&y={a \over c-2b}x-{ab \over c-2b}\\{\overline {CA}}&y={a \over c}x&{\overline {CF}}&y={2a \over 2c-b}x-{ab \over 2c-b}\end{array}}}$

${\displaystyle {\begin{array}{rcl}y&=&{a \over c+b}x\\-[y&=&{a \over c-2b}x-{ab \over c-2b}]\\0&=&{ac-2ab-ac-ab \over (c+b)(c-2b)}x+{ab \over c-2b}\\0&=&{-3ab \over (c+b)(c-2b)}x+{ab \over c-2b}\\{-ab \over c-2b}&=&{-3ab \over (c+b)(c-2b)}x\\1&=&{3 \over c+b}x\\{c+b \over 3}&=&x\\y&=&{a \over c+b}x\\y&=&{\big (}{a \over c+b}{\big )}{\big (}{c+b \over 3}{\big )}\\y&=&{a \over 3}\end{array}}}$

${\displaystyle {\begin{array}{lccl}{\overline {AD}}&{a \over 3}&=&{\big (}{a \over c+b}{\big )}{\big (}{c+b \over 3}{\big )}\\&1&=&1\\{\overline {BE}}&{a \over 3}&=&{a{c+b \over 3}-ab \over c-2b}\\&1&=&{c+b-3b \over c-2b}\\&1&=&{c-2b \over c-2b}\\&1&=&1\\{\overline {CF}}&{a \over 3}&=&{2a{c+b \over 3}-ab \over 2c-b}\\&1&=&{2c+2b-3b \over 2c-b}\\&1&=&{2c-b \over 2c-b}\\&1&=&1\end{array}}}$

Thus, ${\displaystyle {\overline {AD}},\,{\overline {BE}},\,and\,\,{\overline {CF}}}$ pass through point ${\displaystyle {\big (}{\tfrac {c+b}{3}},{\tfrac {a}{3}}{\big )}}$, point ${\displaystyle G}$.


Area Sum
${\displaystyle {\begin{array}{lcllcl}\Delta AEG&=&\int _{0}^{c \over 2}{a \over c}x-{a \over c+b}x\,dx+\int _{c \over 2}^{c+b \over 3}{a \over c-2b}x-{ab \over c-2b}-{a \over c+b}x\,dx&\Delta AFG&=&\int _{0}^{b \over 2}{a \over c+b}x\,dx+\int _{b \over 2}^{c+b \over 3}{a \over c+b}x-{\big (}{2a \over 2c-b}x-{ab \over 2c-b}{\big )}\,dx\\\Delta BDG&=&\int _{c+b \over 3}^{c+b \over 2}{a \over c+b}x-{\big (}{a \over c-2b}x-{ab \over c-2b}{\big )}\,dx+\int _{c+b \over 2}^{b}{\big (}{a \over c-b}x-{ab \over c-b}{\big )}-{\big (}{a \over c-2b}x-{ab \over c-2b}{\big )}\,dx&\Delta BFG&=&\int _{b \over 2}^{c+b \over 3}{2a \over 2c-b}x-{ab \over 2c-b}\,dx+\int _{c+b \over 3}^{b}{a \over c-2b}x-{ab \over c-2b}\,dx\\\Delta CDG&=&\int _{c+b \over 3}^{c}{\big (}{2a \over 2c-b}x-{ab \over 2c-b}{\big )}-{a \over c+b}x\,dx+\int _{c}^{c+b \over 2}{\big (}{a \over c-b}x-{ab \over c-b}{\big )}-{a \over c+b}x\,dx&\Delta CEG&=&\int _{c \over 2}^{c+b \over 3}{a \over c}x-{\big (}{a \over c-2b}x-{ab \over c-2b}{\big )}\,dx+\int _{c+b \over 3}^{c}{a \over c}x-{\big (}{2a \over 2c-b}x-{ab \over 2c-b}{\big )}\,dx\\\Delta ABC&=&\int _{0}^{c}{2a \over c+b}x\,dx+\int _{c}^{b}{a \over c-b}x-{ab \over c-b}\,dx\end{array}}}$

Failed to parse (syntax error): {\displaystyle begin{array}{rcl} \int_0^c {2a\over c+b}x \, dx + \int_c^b {a\over c-b}x-{ab\over c-b} \, dx & = & \int_0^{c\over 2} {a\over c}x-{a\over c+b}x \, dx + \int_{c\over 2}^{c+b\over 3} {a\over c-2b}x-{ab\over c-2b}-{a\over c+b}x \, dx + \int_0^{b\over 2} {a\over c+b}x \, dx + \int_{b\over 2}^{c+b\over 3} {a\over c+b}x-\big({2a\over 2c-b}x-{ab\over 2c-b}\big) \, dx + \int_{c+b\over 3}^{c+b\over 2} {a\over c+b}x-\big({a\over c-2b}x-{ab\over c-2b}\big) \, dx + \int_{c+b\over 2}^b \big({a\over c-b}x-{ab\over c-b}\big)-\big({a\over c-2b}x-{ab\over c-2b}\big) \, dx + \int_{b\over 2}^{c+b\over 3} {2a\over 2c-b}x-{ab\over 2c-b} \, dx + \int_{c+b\over 3}^b {a\over c-2b}x-{ab\over c-2b} \, dx + \int_{c+b\over 3}^c \big({2a\over 2c-b}x-{ab\over 2c-b}\big)-{a\over c+b}x \, dx + \int_c^{c+b\over 2} \big({a\over c-b}x-{ab\over c-b}\big)-{a\over c+b}x \, dx + \int_{c\over 2}^{c+b\over 3} {a\over c}x-\big({a\over c-2b}x-{ab\over c-2b}\big) \, dx + \int_{c+b\over 3}^c {a\over c}x-\big({2a\over 2c-b}x-{ab\over 2c-b}\big) \, dx \\ & = & \int_0^{c\over 2} {a\over c}x-{a\over c+b}x \, dx + \int_{c\over 2}^{c+b\over 3} {a\over c+2b}x-{a\over c+b}x \, dx + \int_{c+b\over 3}^c {a\over c+b}-{a\over c+b}x \, dx + \int_0^{b\over 2} {a\over c+b}x \, dx + \int_{b\over 2}^{c+b\over 3} {a\over c+b}x \, dx + \int_{c+b\over 3}^c \big({2a\over 2c-b}x-{ab\over 2c-b}\big)-{a\over c+b}x \, dx + \int_{c+b\over 3}^c {a\over c+b}x-\big({a\over c-2b}x-{ab\over c-2b}\big) \, dx + \int_c^{c+b\over 2} {a\over c+b}x-\big({a\over c-2b}x-{ab\over c-2b}\big) \, dx + \int_{c+b\over 2}^b \big({a\over c-b}x-{ab\over c-b}\big)-\big({a\over c-2b}x-{ab\over c-2b}\big) \, dx + int_{c+b\over 3}^c {a\over c-2b}x-{ab\over c-2b} \, dx + int_c^b {a\over c-2b}x-{ab\over c-2b} \, dx end{array}}