In △MNO, MN=6units, MO=4 units and NO=12 units. If the bisector of the angle M meets NO at P, calculate NP.


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Question Number 198054 by necx122 last updated on 09/Oct/23

$I n △ M N O, M N = 6 u n i t s, M O = 4 u n i t s$ $a n d N O = 12 u n i t s . I f t h e b i s e c t o r o f t h e$ $a n g l e M m e e t s N O a t P, c a l c u l a t e N P .$
Commented by som(math1967) last updated on 09/Oct/23

$B u t M N + M O = 10$ $N O = 12$ $M N + M O < N O$ $T r i a n g l e p o s s i b l e ?$
Commented by necx122 last updated on 09/Oct/23

$T h e a m s w e r y o u g a v e i n i t i a l l y w a s t h e$ $s a m e t h a t t h e t e x t b o o k g a v e b u t I d i d n t$ $u n d e r s t a n d i t . P l e a s e c a n y o u o r a n y o n e$ $s o l v e w i t h a c l e a r e x p l a n a t i o n . T h a n k s$ $i n a d v a m c e .$
Commented by som(math1967) last updated on 10/Oct/23

$I n △ M N O, b i s e c t o r o f ∠ M$ $c u t N O a t P$ $t h e n \frac{M N}{M O} = \frac{N P}{P O}$ $l e t M N = m, M O = n$ $∴ \frac{N P}{P O} = \frac{m}{n} \Rightarrow N P : P O = m : n$ $i f N O = x u n i t s$ $t h e n N P = x \times \frac{m}{(m + n)} u n i t s$ $P O = x \times \frac{n}{(m + n)} u n i t s$
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