In $\vartriangle ABC$, the external bisectors of $\angle A$ and $\angle B$ meet at a point $D$. Prove that the circumcentre of $\vartriangle ABD$ and the points $C, D$ lie on the same straight line.
Problem
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Tags: geometry, collinear, excenter, Circumcenter
amar_04
11.07.2019 14:49
Join $C$ to $D$. Notice that $CD$ is the angle bisector of $\angle BCA$. Let $(ABC)\cap CD=M$. So, $MB=MA=MI$ where $I$ is the incenter $\triangle ABC$. Now notice that $\angle IAD=90^\circ\implies IM=MA=MD$. So, $MA=MB=MD$. So, $M$ is the circumcenter of $\triangle ABD$. So, $C - M - D$. Hence, proved.
Kamran011
11.07.2019 15:16
This is a direct application of Incenter-Excenter Lemma
akasht
01.06.2020 05:04
Yes but this is an Olympiad for grade 7 to 8 so students aren’t expected to know the Incenter-Excenter Lemma.
yxuann
31.12.2023 05:45
Let O be the circumcenter Let X be the point where CD intersects AB Triangle ACB ≡Triangle ADB(SAS Congruency) Angle AXC= Angle AXD=90 ° Let Y be the Midpoint of the chord of AB in circle Angle AYO=90 °(Angle bisector of chord to centre of circle) Since Angle AXD=Angle AYO=90 °, ∴C,O,D lies on a straight line.
parmenides51
18.05.2024 11:17
Let $I$ be the incenter,
$\angle DAI =90^o= \angle DBI \Rightarrow D,A,B,I$ concyclic with center the midpoint $F$ of $ID$
so $D - F - I$ )( points $D, F, I$ are collinear at this order)
$D$ is $C$-excenter, as intersection of external bisectors of $ \angle A$, $ \angle B$ , so $D$ lies also on the angle bisector of $\angle C$
so $D - I-C$ and because $D - F - I$, we conclude that $D-F-I$