Let $\Delta ABC$ be an acute triangle and $\Omega$ its circumscribed circle, with diameter $AP$. Points $E$ and $F$ are the orthogonal projections from $B$ on $AC$ and $AP$, points $M$ and $N$ are the midpoints of segments $EF$ and $CP$. Prove that $\angle BMN=90$.
Problem
Source: Moldova TST 2020
Tags: geometry
07.03.2020 22:26
What are E, F?
07.03.2020 23:07
07.03.2020 23:16
Notice that the quadrilateral $ABFE$ is cyclic as $\angle{BFA} = \angle{BEA} = 90^{\circ}$. Hence$$\angle{EBF} = \angle{EAF} = \angle{PAC} = \angle{CBP}.$$Also,$$\angle{BEF} = \angle{BAF} = \angle{BAP} = \angle{BCP}.$$Therefore, we can conclude that $\triangle{BEF} \sim \triangle{BCP}$ due to AA similarity. The aforementioned similarity basically tells us that there's a spiral similarity centered at $B$ which sends $\overline{EF}$ to $\overline{CP}$. Now, as $M$ and $N$ are the midpoints of $\overline{EF}$ and $\overline{CP}$, respectively, we can conclude that $M \to N$, which in particular means that $\triangle{BMN} \sim \triangle{BEC}$, and we're done.
08.03.2020 00:04
A complex bash would do too
08.03.2020 00:28
achen29 wrote: What are E, F? Perpendiculars on $AC$ and $AP$
08.03.2020 00:34
achen29 wrote: A complex bash would do too It wouldn't be even so bashy
10.09.2020 15:57
10.09.2020 16:05
?
10.09.2020 16:09
The condition augustin_p wrote: diameter $AP$ is redundant. Even @above's condition is redundant, I think. The same end result follows even if we are given only that $ABCP$ is cyclic.