Extension of angle bisector $BL$ of the triangle $ABC$ (where $AB < BC$) meets its circumcircle at $N$. Let $M$ be the midpoint of $BL$. Isosceles triangle $BDC$ with base $BC$ and angle equal to $ABC$ at $D$ is constructed outside the triangle $ABC$. Prove that $CM \perp DN$. Proposed by А. Mardanov
Problem
Source: Tuymaada 2024 Juniors P6
Tags: geometry
NO_SQUARES
10.07.2024 13:00
Let $S$ be midpoint of $AB$ and $X=CM \cap DN$. Looking on three rotations: with centers $N,D,M$ for which $A \to C \to B \to A$, we get $\angle NSD = 90^\circ$ and $\angle NDS=\angle \frac{1}{2} \angle BDC = \angle SBN$, so $NSBD$ is cyclic. Note that $\angle NBD = \angle NSD = 90^\circ$, so we need to prove that $BMXD$ is cyclic or $180^\circ - \angle BSN = \angle BDN =^? 180^\circ - \angle BMC$ or $\angle BSN =^? \angle BMC$ or (because $\angle NBS = \angle CBM$) $\angle BNS =^? \angle BCM$ or that $CM \cap NS$ lies on $(ABC)$.
Let $K= NS \cap (ABC)$ and $M'=CK \cap BL$. It's enough to prove that $SM' \parallel AC$. Note that $\angle SKM' = \angle NBC = \angle SBM'$, so $BM'SK$ is cyclic. Thus, $\angle SM'K=\angle SBK = \angle ACM'$ and $AC \parallel SM'$, so we are solved the problem!
In lowercase letters, we denote the complex coordinates of the corresponding uppercase ones.
Let $(ABC)$ be a unit circle with $n=-1$, so $ac=1 \Rightarrow a=\frac{1}{c}$. Note that $\Delta (a;1;c) \sim \Delta (b;d;c)$, so \[\frac{1-a}{1-c}=\frac{d-b}{d-c} \Rightarrow d+1=\frac{bc+2c+1}{c+1}.\]
Also $m=(b+ \ell)/2$, there \[\ell = \frac{ac(b+n)-bn(a+c)}{ac-bn}=\frac{b-1+b(c+1/c)}{1+b}=\frac{bc-c+bc^2+b}{c(1+b)}, \]i.e. \[m=\frac{bc+b^2c+bc-c+bc^2+b}{2c(1+b)}\]and thus \[c-m=\frac{2c^2+2c^2b-\left( bc+b^2c+bc-c+bc^2+b) \right)}{2c(1+b)}=\frac{(c-b)(bc+2c+1)}{2c(1+b)}.\]We need to prove that $CM \perp DN$, which is equivalent to \[(c-m)\overline{(d-n)}+\overline{(c-m)}(d-n)=0\]or, putting equalities above, we need to prove that \[ \frac{(bc+2c+1)(b-c)(bc+2b+1)}{2bc(b+1)(c+1)}+\frac{(bc+2b+1)(c-b)(bc+2c+1)}{2bc(b+1)(c+1)}=0\]which is clearly true and we are solved the problem.
sami1618
11.07.2024 17:42
Very interesting geometry : Let $O$ be the circumcenter of $ABC$, let $ON$ meet the circumcircle again at $E$, let $P$ be the reflection of $E$ about $D$. Notice that $E$, $B$, and $D$ are collinear as $\angle DBC=\angle EAC$. The key claim is that triangles $BCL$ and $PEN$ are similar with corresponding pairs of sides perpendicular. This is indeed true as $EN\perp LC$, $EP\perp BL$, and $NP\parallel OD\perp BC$. Thus the corresponding medians are also perpendicular, as desired.
Attachments:
NO_SQUARES
11.07.2024 18:20
Also the following works (in fact, I rewrote official solution): We have \[ DC^2-DM^2=DB^2-DM^2=-BM^2=NB \cdot NL -NM^2=NC^2-NM^2,\]so really $CM \perp DN$.
OutKast
12.07.2024 12:46
Let’s take a circle with Center N and radius NB we know that $I$ and $I_a$ incenter and excepted lies on this circle. $(B,L;I,I_a)=-1$ but M is centre of $BL$ so $MB^2=MI*MI_a$. By angle chasing easy see that $\angle MBD=90$ so MB is tangent to circle with radius DB and Centre D. M lies on radical axis’s of this two circle so we are done
X.Allaberdiyev
04.09.2024 21:49
Radical axes of circles $C$ (imagine that this point is a circle) and the circle with center $M$ and radius $MB$ is $ND$.Thus, $CM \perp DN$
Z4ADies
04.09.2024 23:32
Easy with perpendicularity criterion... We need $DC^2-DM^2=NC^2-NM^2$.We have $DC^2-DM^2=DB^2-DM^2=-BM^2$. $NC$ tangents to $BCL$. So, $NC^2=NL \cdot NB$. Let $NL=x$ and $LM=y$. We need $NL \cdot NB-NM^2=-MC^2$ which is $x(x+2y)-(x+y)^2=-y^2$ and we are done.