# The line through the feet of the bisectors

1. In the previous image $AM$ and $CN$ are bisectors and $P$ lies on the line segment $MN$. Prove that the distance from $P$ to line $AC$ is equal to the sum of the distances from $P$ to lines $BA$ and $BC$.

2. What happen if $P$ lies on the line $MN$ but outside the triangle?

3. Now draw the circumcircle of triangle $ABC$ and suppose the ray $MN$ intersects the circumcircle at $Q$ ($N$ is between $Q$ and $M$). Prove that $\frac{1}{QA}=\frac{1}{QC}+\frac{1}{QB}$.

Note by Jorge Tipe
5 years, 6 months ago

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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 )

- 5 years, 5 months ago

Reply if you find this problem interesting! :)

- 5 years, 5 months ago

I think there are too many conditions in your problem. Triangle ABC is fixed and also the points M and N, thus if you choose P in line segment MN such that [BPN]=2 then [CPM] is determined and is not equal to 4. Another idea is to calculate [MNBC] and then find [BPC] (I think the answer will be different of yours).

Try this problem (using your notation):

If [BPN]=2, what is the correct value of [CMP]?

- 5 years, 5 months ago

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

- 5 years, 5 months ago

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 $PY + PZ = PL + LX = PX$

- 5 years, 5 months ago

Is it correct Jorge??

- 5 years, 5 months ago

It seems good, but you misplaced the points, for example you said $MM_1$ perpendicular to $BC$.

- 5 years, 5 months ago