Question Number 189438 by normans last updated on 16/Mar/23
Answered by som(math1967) last updated on 16/Mar/23
$${construct}\:{FM}\parallel{BE} \\ $$$$\bigtriangleup{AFM}\sim\bigtriangleup{ABE} \\ $$$$\frac{{AM}}{{AE}}=\frac{\mathrm{3}}{\mathrm{3}+\mathrm{5}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\frac{{AM}}{{ME}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${EO}\parallel{FM} \\ $$$$\frac{{CE}}{{EM}}=\frac{{CO}}{{OF}} \\ $$$$\frac{{CE}}{{EA}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\frac{{CE}}{{EA}}×\frac{{AE}}{{AM}}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\frac{{CE}}{{AM}}=\frac{\mathrm{16}}{\mathrm{9}} \\ $$$$\frac{{CE}}{{AM}}×\frac{{AM}}{{ME}}=\frac{\mathrm{16}}{\mathrm{9}}×\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\frac{{CE}}{{EM}}=\frac{\mathrm{16}}{\mathrm{15}}=\frac{{CO}}{{OF}} \\ $$$$\therefore\frac{{ar}\bigtriangleup{BFO}}{{ar}\bigtriangleup{BCO}}=\frac{{FO}}{{CO}}=\frac{\mathrm{15}}{\mathrm{16}} \\ $$
Commented by som(math1967) last updated on 16/Mar/23
