Let $ABCD$ be a square of center $O$ and let $M$ be the symmetric of the point $B$ with respect to point $A$. Let $E$ be the intersection of $CM$ and $BD$, and let $S$ be the intersection of $MO$ and $AE$. Show that $SO$ is the angle bisector of $\angle ESB$.
Problem
Source: BMO SL 2019, G1
Tags: geometry, Angle Chasing, bisect, square
08.11.2020 04:36
Seems approachable by Coordinates
Attachments:

09.11.2020 20:42
Properties of this configuration : $ASOB$ is cyclic, as well as $MDSA$ $AO$ is tangent to $(MAS)$ and $(DAS)$ Triangles $ASO$ and $ASD$ are similar, as well as $MAO$ and $MAC$
09.11.2020 22:37
Assume that $MC \cap AD = X$. It is easy to see that $AX$ is midline in the $\triangle MBC$, therefore $X$ is midpoint of $AD$ and $OX \parallel MB$. Claim: Points $B,S,X$ are collinear. Proof: Assume that $BS \cap MC = X'$. By Ceva's theorem in $\triangle MBE$, we have that $\frac{OE}{BO} =\frac{EX'}{X'M}$. This forces to have that $OX' \parallel MB$, which implies that $X = X'$ Let $Y$ be midpoint of $CD$. It is easy to see that $G$ is centroid of $\triangle ACD$, therefore points $A,S,E,Y$ are collinear. Observe that $\triangle BAX = \triangle DAY$. Simple angle chasing reveals that $\angle BSE = 90^{\circ}$. Since we have that $\angle BOA = \angle BSA =90^{\circ}$, then quadrilateral $ABOS$ is cyclic and consequently we have that $\angle BAO =\angle BSO = \angle OSE =45^{\circ}$. This proves that $SO$ is angle bisector of $\angle BSE$ as desired.
25.02.2022 21:04
Let $BS$ meet $MC$ at $N$. Claim1 : $N$ is intersection of $AD$ and $MC$. Proof : By Ceva Theorem we have $\frac {MA}{AB} . \frac {BO}{OE} . \frac {EN}{NM} = 1$ so $\frac {BO}{OE} = \frac {NM}{EN}$ so $ON || MB$. Now we have $\angle EAD = \angle ECD = \angle NMA = \angle NBA$ so we have $\angle BSA = \angle 90 = \angle BOA$ so $ASOB$ is cyclic and we have $OA = OB$ so $\angle ASM = \angle BSO$ so $\angle ESO = \angle OSB$. we're Done.
29.08.2022 22:43
Barycentric coordinates: $A(1,0,0),B(0,1,0),C(0,0,1),O(1:0:1),M(2,-1,0),$ where $a=c,a^2+c^2=b^2$ Calculate trivial determinants:$E(2:-1:2)$ , $S(4:-1:2)$. One method is calculating the ratio of sides $\frac{SE}{SB}=\frac{EO}{OB}$. Or, using circle equation formula,$(AOB)=(ABS)\Rightarrow$ $AOSB$ is cyclic. By cyclic, $90=\angle BOA= \angle BSA \Rightarrow \angle BSE=90$ and $45=\angle BAC= \angle BSO \rightarrow \angle BSO=45$. Since $\angle BSE-\angle BSO=45=\angle BSO$, we are done $\blacksquare$
30.08.2022 08:45
Dear Mathlinkers, https://artofproblemsolving.com/community/c6t48f6h2334497_square_geometry_bisect_angle_esb Sincerely Jean-Louis
28.01.2025 15:58
λυση αργοτερα
28.01.2025 20:47
Let one side of the square, ie. $AB=6x$. Quick calculation gives $MS=\dfrac{12x\sqrt{10}}{5}$ and $MO=3x\sqrt{10}$. Hence, $$MA\cdot MB=72x^2=MS\cdot MO$$which implies $ASOB$ is cyclic, $AS\perp BS$. And $\angle OSB=45^{\circ}$ giving $\angle ESO=\angle BSO$ as desired.
28.01.2025 21:06
VicKmath7 wrote: Properties of this configuration : $ASOB$ is cyclic, as well as $MDSA$ $AO$ is tangent to $(MAS)$ and $(DAS)$ Triangles $ASO$ and $ASD$ are similar, as well as $MAO$ and $MAC$ It has been proved above that $ASOB$ cyclic. Claim: MASD is cyclic. PROOF. We know $MS=12x\sqrt{10}/5$ and $AP=2x$. Thus $MP=2x\sqrt{10}$ and $PS=2x\sqrt{10}/5$. Hence $MP\cdot PS=8x^2=AP\cdot PF$ holds, implying $MASD$ is cyclic. Claim: $AO$ is tangent to $(MAS)$ and $(DAS)$. PROOF. Let $AO\cap SB=K$. Then $$\angle AMS=45^{\circ}-\angle AOM=90^{\circ}-\angle AKS=\angle SAO$$proving our claim. $\angle AMO=\angle ADO$ shows the second tangency as well. Claim: $ASD\sim OSA$. PROOF. We showed in tangency that $\angle ADS=\angle SAO$. It sufficies to show $\angle DAS=\angle AOS$. On the other hand, $$\angle DAS=45^{\circ}-\angle SAO=45^{\circ}+\angle AOS-45^{\circ}=\angle AOS$$giving the similarity between triangles. Claim: $MAO\sim CAM$ PROOF. $MA^2=AO\cdot AC$ implies $(MOC)$ is tangent to $MA$ at $M$ giving the similarity.