Given a triangle $ABC$, in which $AB = BC$. Point $O$ is the center of the circumcircle, point $I$ is the center of the incircle. Point $D$ lies on the side $BC$, such that the lines $DI$ and $AB$ parallel. Prove that the lines $DO$ and $CI$ are perpendicular.
(Vyacheslav Yasinsky)
We have $\angle IOC=180^\circ-\angle COB=\beta=\angle IDC$, so $IOCD$ is cyclic. But then $\angle ODC+\angle DCI=180^\circ-\angle CIO+\frac{\gamma}{2}=\frac{\beta}{2}+\gamma=90^\circ$. Done.
Tintarn wrote:
We have $\angle IOC=180^\circ-\angle COI=\beta=\angle IDC$, so $IOCD$ is cyclic. But then $\angle ODC+\angle DCI=180^\circ-\angle CIO+\frac{\gamma}{2}=\frac{\beta}{2}+\gamma=90^\circ$. Done.
But $<IOC=<COI$
Tintarn wrote:
Sorry it was a typo. It should be $\angle IOC=\angle COB$. But I think it was clear what is meant, anyway.
Yes is was clear. but still I want to make sure