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Question Number 23262 by ajfour last updated on 28/Oct/17

Commented by ajfour last updated on 28/Oct/17

((CN)/(AC))=(x/(x+y))   ;   ((AN)/(AC))=(y/(x+y))  ((MB)/(AB)) =(y/(x+y)) ;  ((AM)/(AB))=(x/(x+y)) .  based on similarity  of triangles.  △CNP ∼ △CAB  ⇒  ((CN)/(AC))=((NP(=AM))/(AB))=((CP)/(BC))   If AB=a  and AC=b , then  ⇒  ((CN)/b)=((AM)/a)=(x/(x+y))  ⇒  CN=((x/(x+y)))b ;  AM=((x/(x+y)))a   AN=AC−CN =b−((bx)/(x+y))=((y/(x+y)))b   MB=AB−AM=a−((ax)/(x+y))=((y/(x+y)))a .

$$\frac{{CN}}{{AC}}=\frac{{x}}{{x}+{y}}\:\:\:;\:\:\:\frac{{AN}}{{AC}}=\frac{{y}}{{x}+{y}} \\ $$$$\frac{{MB}}{{AB}}\:=\frac{{y}}{{x}+{y}}\:;\:\:\frac{{AM}}{{AB}}=\frac{{x}}{{x}+{y}}\:. \\ $$$${based}\:{on}\:{similarity}\:\:{of}\:{triangles}. \\ $$$$\bigtriangleup{CNP}\:\sim\:\bigtriangleup{CAB} \\ $$$$\Rightarrow\:\:\frac{{CN}}{{AC}}=\frac{{NP}\left(={AM}\right)}{{AB}}=\frac{{CP}}{{BC}} \\ $$$$\:{If}\:{AB}={a}\:\:{and}\:{AC}={b}\:,\:{then} \\ $$$$\Rightarrow\:\:\frac{{CN}}{{b}}=\frac{{AM}}{{a}}=\frac{{x}}{{x}+{y}} \\ $$$$\Rightarrow\:\:{CN}=\left(\frac{{x}}{{x}+{y}}\right){b}\:;\:\:{AM}=\left(\frac{{x}}{{x}+{y}}\right){a} \\ $$$$\:{AN}={AC}−{CN}\:={b}−\frac{{bx}}{{x}+{y}}=\left(\frac{{y}}{{x}+{y}}\right){b} \\ $$$$\:{MB}={AB}−{AM}={a}−\frac{{ax}}{{x}+{y}}=\left(\frac{{y}}{{x}+{y}}\right){a}\:. \\ $$

Commented by ajfour last updated on 28/Oct/17

Explanation to a query related to  Q. 23251

$${Explanation}\:{to}\:{a}\:{query}\:{related}\:{to} \\ $$$${Q}.\:\mathrm{23251} \\ $$

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