In-ABC-the-following-relationship-holds-m-b-b-m-c-c-a-2r-n-b-h-b-n-c-h-c-

Question Number 183167 by Shrinava last updated on 21/Dec/22

In △ ABC the following relationship

holds :

\frac{m_{b}}{b} + \frac{m_{c}}{c} ⩽ \frac{a}{2 r} ⩽ \frac{n_{b}}{h_{b}} + \frac{n_{c}}{h_{c}}

Commented by mr W last updated on 22/Dec/22

i t h i n k n o t e v e r y b o d y k n o w s w h a t i s

w h a t i n t h e e q u a t i o n .

Commented by Frix last updated on 22/Dec/22

{That}^{'} s plain to see because obviously

(\frac{α}{ρ_{b} + ρ_{c}} + \frac{β}{ρ_{a} + ρ_{c}}) (i_{c} - j_{c}) > \frac{γ}{ρ_{a} + ρ_{b}} ((i_{a} - j_{a}) + (i_{b} - j_{b}))

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