# Triangle and reflections

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.

• 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.) Aug 25, 2020 at 14:59
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 $$A$$.