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Question Number 21571 by Tinkutara last updated on 27/Sep/17

ABCD is a cyclic quadrilateral; x, y, z are the distances of A from the lines BD, BC, CD respectively. Prove that ((BD)/x) = ((BC)/y) + ((CD)/z).

$${ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cyclic}\:\mathrm{quadrilateral};\:{x},\:{y},\:{z} \\ $$$$\mathrm{are}\:\mathrm{the}\:\mathrm{distances}\:\mathrm{of}\:{A}\:\mathrm{from}\:\mathrm{the}\:\mathrm{lines} \\ $$$${BD},\:{BC},\:{CD}\:\mathrm{respectively}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{{BD}}{{x}}\:=\:\frac{{BC}}{{y}}\:+\:\frac{{CD}}{{z}}. \\ $$

Answered by revenge last updated on 29/Sep/17

Commented by revenge last updated on 29/Sep/17

It can be proved that: (i) ΔADE ∼ ΔABG (ii) ΔADF ∼ ΔACG (iii) ΔAFB ∼ ΔAEC From (i), ((BG)/y)=((DE)/z) ⇒ ((BC)/y)+((CD)/z)=((GC−GB)/y)+((CE+ED)/z)=((GC)/y)+((CE)/z) ...(∗) From (ii), ((CG)/y)=((DF)/x) ...(1) From (iii), ((CE)/z)=((FB)/x) ...(2) Adding (1) & (2) with (∗), we get ((BC)/y)+((CD)/z)=((DF+FB)/x)=((BD)/x)

$${It}\:{can}\:{be}\:{proved}\:{that}: \\ $$$$\left({i}\right)\:\Delta{ADE}\:\sim\:\Delta{ABG} \\ $$$$\left({ii}\right)\:\Delta{ADF}\:\sim\:\Delta{ACG} \\ $$$$\left({iii}\right)\:\Delta{AFB}\:\sim\:\Delta{AEC} \\ $$$${From}\:\left({i}\right),\:\frac{{BG}}{{y}}=\frac{{DE}}{{z}} \\ $$$$\Rightarrow\:\frac{{BC}}{{y}}+\frac{{CD}}{{z}}=\frac{{GC}−{GB}}{{y}}+\frac{{CE}+{ED}}{{z}}=\frac{{GC}}{{y}}+\frac{{CE}}{{z}}\:...\left(\ast\right) \\ $$$${From}\:\left({ii}\right),\:\frac{{CG}}{{y}}=\frac{{DF}}{{x}}\:...\left(\mathrm{1}\right) \\ $$$${From}\:\left({iii}\right),\:\frac{{CE}}{{z}}=\frac{{FB}}{{x}}\:...\left(\mathrm{2}\right) \\ $$$${Adding}\:\left(\mathrm{1}\right)\:\&\:\left(\mathrm{2}\right)\:{with}\:\left(\ast\right),\:{we}\:{get} \\ $$$$\frac{{BC}}{{y}}+\frac{{CD}}{{z}}=\frac{{DF}+{FB}}{{x}}=\frac{{BD}}{{x}} \\ $$

Commented by Tinkutara last updated on 29/Sep/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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