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# 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
3 years, 2 months ago

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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 $$PY + PZ = PL + LX = PX$$ · 3 years, 2 months ago

Is it correct Jorge?? · 3 years, 2 months ago

It seems good, but you misplaced the points, for example you said $$MM_1$$ perpendicular to $$BC$$. · 3 years, 2 months ago

Then cant we fix $$P$$ and say that suppose there exists a point $$P$$ on $$MN$$ such that ........ ? · 3 years, 2 months ago

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 ) · 3 years, 2 months ago

Reply if you find this problem interesting! :) · 3 years, 2 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]? · 3 years, 2 months ago