Question Number 21571 by Tinkutara last updated on 27/Sep/17
$${ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cyclic}\:\mathrm{quadrilateral};\:{x},\:{y},\:{z} \\ $$$$\mathrm{are}\:\mathrm{the}\:\mathrm{distances}\:\mathrm{of}\:{A}\:\mathrm{from}\:\mathrm{the}\:\mathrm{lines} \\ $$$${BD},\:{BC},\:{CD}\:\mathrm{respectively}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{{BD}}{{x}}\:=\:\frac{{BC}}{{y}}\:+\:\frac{{CD}}{{z}}. \\ $$
Answered by revenge last updated on 29/Sep/17
Commented by revenge last updated on 29/Sep/17
$${It}\:{can}\:{be}\:{proved}\:{that}: \\ $$$$\left({i}\right)\:\Delta{ADE}\:\sim\:\Delta{ABG} \\ $$$$\left({ii}\right)\:\Delta{ADF}\:\sim\:\Delta{ACG} \\ $$$$\left({iii}\right)\:\Delta{AFB}\:\sim\:\Delta{AEC} \\ $$$${From}\:\left({i}\right),\:\frac{{BG}}{{y}}=\frac{{DE}}{{z}} \\ $$$$\Rightarrow\:\frac{{BC}}{{y}}+\frac{{CD}}{{z}}=\frac{{GC}−{GB}}{{y}}+\frac{{CE}+{ED}}{{z}}=\frac{{GC}}{{y}}+\frac{{CE}}{{z}}\:…\left(\ast\right) \\ $$$${From}\:\left({ii}\right),\:\frac{{CG}}{{y}}=\frac{{DF}}{{x}}\:…\left(\mathrm{1}\right) \\ $$$${From}\:\left({iii}\right),\:\frac{{CE}}{{z}}=\frac{{FB}}{{x}}\:…\left(\mathrm{2}\right) \\ $$$${Adding}\:\left(\mathrm{1}\right)\:\&\:\left(\mathrm{2}\right)\:{with}\:\left(\ast\right),\:{we}\:{get} \\ $$$$\frac{{BC}}{{y}}+\frac{{CD}}{{z}}=\frac{{DF}+{FB}}{{x}}=\frac{{BD}}{{x}} \\ $$
Commented by Tinkutara last updated on 29/Sep/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$