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:
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!
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 $A$.