$2.$ Let $ABC$ be a triangle and $AD$ the angle bisector ($D\in BC$). The perpendicular from $B$ to $AD$ cuts the circumcircle of triangle $ABD$ at $E$. If $O$ is the center of the circle around $ABC$ , prove $A,O,E$ are collinear.
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Tags: geometry
Mr.Mister
09.10.2019 20:12
Let BAC = x ; ABC = y => BAD = x/2 => ABE = 90 - x/2 => DBE = y + x/2 - 90 = DAE. we know OAC = 90 - y => DAO = x - x/2 - (90 - y) = y + x/2 - 90 thus A, O, E lie on the same line.
itslumi
27.11.2019 19:26
Hint ;just drop the altitude from A.
IvanMitrovic
05.02.2020 14:10
So let F be a pependicular from B to AD. Now we have that <EBA=<FBA=90°-<BAC/2. Since <AEB=<ADB=<ACB+<BAC/2 we have that <EAB=180°-<AEB-<EBA=180°-<ACB-<BAC/2-90°+<BAC/2=90°-<ACB. On the other side we have that <AOB=2<ACB and <OAB=90°-<ACB. Finally <OAB=<EAB and A, O and E are collinear.
rafigamath
25.05.2023 16:00
Let $\angle{ABE}=\alpha , \angle{EBD}=\beta$. Since $\angle{AOC}$ is the central angle $\angle{OAC}= {90- \alpha-\beta}$. $\angle{BAD}=\angle{DAC}={90-\alpha}$ and $\angle{EAD}= {\beta}$ $\implies \angle{EAC}={90-\alpha-\beta}$ and we're done
tooral888
29.05.2023 17:28
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