Prove that $(A,N;P,B)=(A,M;Q,C)=-1$ .

Question

Given $\Delta ABC$ and incircle $\omega$ tangent to $BC,AC,AB$ at $Y,M,N$ respectively . Let $AY \cap \omega=X$ . Let the tangent through $X$ wrt $\omega$ intersect $AB$ and $AC$ at $P$ and $Q$ respectively . Prove that $(A,N;P,B)=(A,M;Q,C)=-1$ .

I am completely stuck in this problem. I tried considering the intersection of $FE$ and $BC$, but no movement .

I didn't understand the meaning of the notations you used in "To prove...". Can you please explain? — SarGe, Aug 15 '20 at 08:00
@MichaelRozenberg yes, the directed angles are coming out as positive .. I edited the question , it's actually a lemma which I need to prove .. — Sunaina Pati, Aug 15 '20 at 08:10
@Shubhangi, ok, got it. I think that proving your statement is equivalent to proving that "$B$ is harmonic conjugate of $P$ w.r.t. $A$ and $N$. — SarGe, Aug 15 '20 at 08:13
Cross ratio = $-1$ is a definition of harmonic conjugate of a triple of points. So, you just have to show that $P$ divides $AN$ in the same ratio internally, as $B$ does externally. — Fawkes4494d3, Aug 15 '20 at 08:15
By Brianchon, $BQ, CP, XY, MN$ are concurrent. Can you finish it now? — timon92, Aug 15 '20 at 08:21
Just to add timon's point more, notice if you let $MN \cap AY=J$ then $(AJ,XY)=-1$ — , Aug 15 '20 at 08:27
@timon92 I am very new to Projective Geo, Can you if possible state the statement of Brianchon ? — Sunaina Pati, Aug 15 '20 at 08:50
@Shubhangi You can see here https://en.wikipedia.org/wiki/Brianchon%27s_theorem — Anand, Aug 15 '20 at 09:02

score 3 · Answer 1 · answered Aug 15 '20 at 12:55

If you familiar with harmonic quadrilaterals, then this can be proved in the following way.

Let $T$ be the intersection of the incircle and line $YP$ (and $Y\neq T$). Then, the quadrilateral $YNTX$ is harmonic. Indeed, it's cyclic and tangents at points $X$ and $Y$ intersect at point $P$ on line $YT$. Thus, $$ (Y,T;N,X)=-1. $$ Now, note that $$ (Y,T;N,X)=(B,P;N,A), $$ because we can project the incircle onto line $AB$ from point $Y$ (lines $YY$, $YT$, $YN$ and $YX$ intersect line $AB$ at points $B$, $P$, $N$ and $A$, respectively). Therefore, $$ (B,P;N,A)=-1,~\text{so}~(A,N;B,P)=-1, $$ as desired. Similarly, we can show that $(A,M;Q,C)=-1$.

Here, line $YY$ is the tangent to the incircle at point $Y$. — richrow, Aug 15 '20 at 12:57

Anand · Answer 2 · 2020-08-15T11:22:13.487

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It's well known that $BQ, CP, XY, NM$ concur say at $O$. Taking perspectivity at $O$, we get $$(AM;QC)\overset{O}{=}(AN;PB)$$ To show that this is a harmonic bundle note that $$(AN;PB)=-1\implies \frac{AP}{NP}=\frac{AB}{NB}\iff \frac{AP}{PX}=\frac{AB}{BY}$$ Now as $\angle PXY =\angle BYX$, we get $\frac{AP}{PX}=\frac{AB}{BY}$ by ratio lemma. Completeing the proof.

edited Aug 15 '20 at 11:22

answered Aug 15 '20 at 08:52

Anand

3,588

well, how do you get the last part ? – Sunaina Pati Aug 15 '20 at 10:04
@Shubhangi That's just converse of Thales Theorem – Anand Aug 15 '20 at 10:17
i see.. .. I must have understood the question wrongly..Thanks! – Sunaina Pati Aug 15 '20 at 10:34
The last equivalence is incorrect. The problem statement is true by the way, I'll post an elementary solution in a sec. – timon92 Aug 15 '20 at 10:42
@timon92 Yeah, I fixed it – Anand Aug 15 '20 at 11:22
@Anand how did you get ", we get /=/" ? – Sunaina Pati Aug 15 '20 at 15:20
@Shubhangi Just use ratio lemma (basically sine rule) in triangle $APX$ and $ABY$ by noting that $\sin \angle AXP = \sin \angle AYB$. – Anand Aug 15 '20 at 17:30
@Anand Thanks , I got it ... BTW congrats for SOR !! – Sunaina Pati Aug 16 '20 at 12:20
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@Shubhangi Thank you :)...How do you know bout SOR? – Anand Aug 16 '20 at 12:23

score 2 · Accepted Answer · answered Aug 15 '20 at 10:47

Let $Z$ be a point on $AY$, different from $X$, such that $PX=PZ$. Then $\angle XZP = \angle PXZ = \angle XYB$, hence $PZ \parallel BY$. Hence $\dfrac{AP}{PZ} = \dfrac{AB}{BY}$. Since $PZ=PX=PN$ and $BY=BN$, we obtain $\dfrac{AP}{PN} = \dfrac{AB}{BN}$ which is equivalent to $(A,N;P,B)=-1$.

Prove that $(A,N;P,B)=(A,M;Q,C)=-1$ .

3 Answers3