The line through the feet of the bisectors

I proved The first 1... and based on that created a problem. Maybe if u likeJorge,,,, you may share this problem. So here's the problem. In an acute angled triangle $ABC$ we have $BC=4$,$AC=5$,$AB=6$. $BM$ and $CN$ are the internal bisectors of $B$ and $C$. $P$ is any point on the segment $MN$. If $[\triangle PMC]=1$ and $[\triangle PNB]=4$ , find $[\triangle BPC]$ . ( it wud be nice if u mention a little credit to me.. :P )

Then cant we fix $P$ and say that suppose there exists a point $P$ on $MN$ such that ........ ?

However, solution to #1. Let $PX \perp BC$, $PY \perp AC$ and $PZ \perp AB$. We drop $MM_1 \perp BC$, $MZ_1 \perp AB$ and $NN_1 \perp BC$ and $NY_1 \perp AC$. Now we have $\triangle BM_1M \cong \triangle BMZ_1$ (right angle, angle and side equal) and hence $MZ_1=MM_1$. Similarly we have, $NN_1 = NY_1$.

We use similarity in triangles $MPY$ and $MNY_1$.

$\dfrac{PY}{NY_1} = \dfrac{MP}{NM}$ and we get $PY = \dfrac{PM}{NM} \times NY_1$...(i) Similarly we have $PZ = \dfrac{NP}{NM} \times MZ_1$...(ii)

Now, $NN_1 , PX,MM_1$ are 3 parallel lines . So we have $\dfrac{PM}{NM} = \dfrac{M_1X}{M_1N_1}$.... (iii) We join $N, M_1$ and let that intersect $PX$ at $L$.

Using $\triangle NLP \sim NM_1M$ , we have $\dfrac{PL}{MM_1} = \dfrac{NP}{NM}$ and thus $PL = \dfrac{NP}{NM} \times MM_1 = \dfrac{NP}{NM} \times MZ_1$. (as $MM_1 = MZ_1$ ... (iv)

And, similarly using $\triangle M_1XL \sim \triangle M_1N_1N$ , we have

$LX = \dfrac{XM_1}{N_1M_1} \times NN_1 = \dfrac{PM}{NM} \times NY_1$...(v)

We observe that $(iv) + (v)$ is indeed $(i) + (ii)$ and Hence $PY + PZ = PL + LX = PX$