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Let $ABC$ be a triangle where all angles measure less than $90$ degrees. $M$ and $N$ are 2 points on $BC$, in this order. $P$ and $Q$ are the reflections of $M$ and $N$ on $AB$, while $S$ and $R$ are the reflections of $M$ and $N$ on $AC$.

I have to prove the following things:

a)$[PQ]\equiv[SR]$

b)$RC$ and $BP$ intersect in a point $T$ and $A$ and $T$ are on different sides of $BC$

c)$[TA$ is the bisector of $\angle RTP$

I have made a drawing in Geogebra and marked some extra points: Geogebra Drawing

a)The triangles $EFP$ and $EFM$ are congruent, therefore triangles $QEP$ and $NEP$ are congruent, so $PQ\equiv MN$. Similarly $RS\equiv MN$, therefore $PQ\equiv SR$

b)Triangles $FBP$ and $FBM$ are congruent, therefore angles $PBF$ and $ABC$ are congruent. Similarly angles $HCR$ and $ACB$ are congruent. Therefore:

$$m(\angle PBC)+m(\angle BCR)=2(m(\angle ABC)+m(\angle ACB))=2(180-m(\angle ABC)$$

The right side is greater than $180$ degrees(because $ABC$'s angles are all less than $90$ degrees). Therefore $PB$ and $RC$ intersect, and they do so "under" $BC$.

c)This is were I pretty much got stuck. I proved through some triangle congruences that $Q$, $P$, $B$ and $S$, $R$, $C$ are collinear. I also found that $m(\angle BTC)=180-2m(\angle ABC)$. I don't know whether these help with anything.

I suppose this could be solved by adding a coordinates system and bashing calculations, but I don't think that's how the problem is supposed to be solved.

Thanks for your help!

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    $\begingroup$ As per your image, $P$ and $Q$ are reflections of $M,N$ with respect to $AB$, not projections onto it, and similar with $R, S$. (With the word "projections", $P$ would coincide with $F$, $Q$ would coincide with $E$ etc. - and the problem won't make sense.) $\endgroup$ Aug 25, 2020 at 14:59
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    $\begingroup$ Thanks, I have edited it. $\endgroup$
    – Wolfuryo
    Aug 25, 2020 at 15:03

1 Answer 1

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In $\triangle BCT$, $BA$ and $CA$ are the bisectors of the outer angles $\angle B$ and $\angle C$, so their intersection $A$ is the centre of the "excircle" (the circle touching all three sides of the triangle - one of them from the "outside"). Thus, the third bisector (of the angle $\angle T$) must also go through $A$.

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