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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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 \(PY + PZ = PL + LX = PX\)