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Question Number 67973 by behi83417@gmail.com last updated on 02/Sep/19

Commented by behi83417@gmail.com last updated on 02/Sep/19

AB^△ C,is given. show that: 1. ((BX)/(XC)) × ((CY)/(YA)) × ((AZ)/(ZB))=1. 2. ((PX)/(XA)) + ((PY)/(YB)) + ((PZ)/(ZC))=1.

$$\mathrm{A}\overset{\bigtriangleup} {\mathrm{B}C},\mathrm{is}\:\mathrm{given}. \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}: \\ $$$$\mathrm{1}.\:\:\:\:\:\frac{\boldsymbol{\mathrm{BX}}}{\boldsymbol{\mathrm{XC}}}\:×\:\:\frac{\boldsymbol{\mathrm{CY}}}{\boldsymbol{\mathrm{YA}}}\:×\:\frac{\boldsymbol{\mathrm{AZ}}}{\boldsymbol{\mathrm{ZB}}}=\mathrm{1}. \\ $$$$\mathrm{2}.\:\:\:\:\:\frac{\boldsymbol{\mathrm{PX}}}{\boldsymbol{\mathrm{XA}}}\:+\:\:\frac{\boldsymbol{\mathrm{PY}}}{\boldsymbol{\mathrm{YB}}}\:+\:\frac{\boldsymbol{\mathrm{PZ}}}{\boldsymbol{\mathrm{ZC}}}=\mathrm{1}. \\ $$

Commented by Tony Lin last updated on 03/Sep/19

1. { ((((△ABP)/(△ACP))=((BX)/(XC))(PA is the common bottom edge))),((((△BCP)/(△ABP))=((CY)/(YA))(PB is the common bottom edge))),((((△ACP)/(△BCP))=((AZ)/(ZB))(PC is the common bottom edge))) :} ∴((△BCP)/(△ABP))×((△ABP)/(△ACP))×((△ACP)/(△BCP)) =((BX)/(XC))×((CY)/(YA))×((AZ)/(ZB)) =1

$$\mathrm{1}.\begin{cases}{\frac{\bigtriangleup{ABP}}{\bigtriangleup{ACP}}=\frac{{BX}}{{XC}}\left({PA}\:{is}\:{the}\:{common}\:{bottom}\:{edge}\right)}\\{\frac{\bigtriangleup{BCP}}{\bigtriangleup{ABP}}=\frac{{CY}}{{YA}}\left({PB}\:{is}\:{the}\:{common}\:{bottom}\:{edge}\right)}\\{\frac{\bigtriangleup{ACP}}{\bigtriangleup{BCP}}=\frac{{AZ}}{{ZB}}\left({PC}\:{is}\:{the}\:{common}\:{bottom}\:{edge}\right)}\end{cases} \\ $$$$\therefore\frac{\bigtriangleup{BCP}}{\bigtriangleup{ABP}}×\frac{\bigtriangleup{ABP}}{\bigtriangleup{ACP}}×\frac{\bigtriangleup{ACP}}{\bigtriangleup{BCP}} \\ $$$$=\frac{{BX}}{{XC}}×\frac{{CY}}{{YA}}×\frac{{AZ}}{{ZB}} \\ $$$$=\mathrm{1} \\ $$

Commented by behi83417@gmail.com last updated on 03/Sep/19

thank you very much sir.please try part#2 and Q#67969.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir}.\mathrm{please}\:\mathrm{try} \\ $$$$\mathrm{part}#\mathrm{2}\:\mathrm{and}\:\mathrm{Q}#\mathrm{67969}. \\ $$

Commented by Tony Lin last updated on 03/Sep/19

2. △ABC=△ABP+△BCP+△ACP ((△ABC)/(△ABC))=((△ABP)/(△ABC))+((△BCP)/(△ABC))+((△ACP)/(△ABC)) ⇒((PZ)/(ZC))+((PX)/(XA))+((PY)/(YB))=1

$$\mathrm{2}. \\ $$$$\bigtriangleup{ABC}=\bigtriangleup{ABP}+\bigtriangleup{BCP}+\bigtriangleup{ACP} \\ $$$$\frac{\bigtriangleup{ABC}}{\bigtriangleup{ABC}}=\frac{\bigtriangleup{ABP}}{\bigtriangleup{ABC}}+\frac{\bigtriangleup{BCP}}{\bigtriangleup{ABC}}+\frac{\bigtriangleup{ACP}}{\bigtriangleup{ABC}} \\ $$$$\Rightarrow\frac{{PZ}}{{ZC}}+\frac{{PX}}{{XA}}+\frac{{PY}}{{YB}}=\mathrm{1} \\ $$

Commented by mr W last updated on 03/Sep/19

nice work, Lin sir!

$${nice}\:{work},\:{Lin}\:{sir}! \\ $$

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