The line through the feet of the bisectors

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 $BC$ .
What happen if $P$ lies on the line $MN$ but outside the triangle?
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 $\frac{1}{QA}=\frac{1}{QC}+\frac{1}{QB}$ .

#Geometry #Distance #Circumcircle #PeruMOTraining

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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 )

Sagnik Saha - 6 years ago

Reply if you find this problem interesting! :)

Sagnik Saha - 6 years 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]?

Jorge Tipe - 6 years ago

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

Sagnik Saha - 6 years 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 $PY + PZ = PL + LX = PX$

Sagnik Saha - 6 years ago

Is it correct Jorge??

Sagnik Saha - 6 years ago

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

Jorge Tipe - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$