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If $I$ is the incentre of $\triangle ABC$ and $AI$ meets the circumcircle at $D$, the prove that $DB = DC = DI$.

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Here, I am basically trying to prove that $B$, $I$ and $C$ all lie on a circle centered at $D$. $ABCD$ is a cyclic quadrilateral so it's opposite angles must be supplementary. Also, $\angle BDC = 180 - \angle A$.

I am unable to proceed from here. Please help.

Thanks.

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    $\begingroup$ $\widehat{BAD}=\widehat{DAC}$ implies that the arcs $BD$ and $DC$ have the same length, so they do the chords $BD$ and $DC$. The fact that both $DIB$ and $DIC$ are isosceles triangles follows from angle chasing. $\endgroup$ Oct 2 '16 at 14:49
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Let me try. We have $$\angle CID = \angle ICA + \angle IAC = \frac{\angle A}{2} + \frac{\angle C}{2}$$.

On other hand, we have $$\angle DCI = \angle DCB + \angle BCI = \angle DAB + \angle BCI = \frac{\angle A}{2} + \frac{\angle C}{2}.$$

Then, you have $\Delta DIC$ is an isosceles triangle, or $DI = DC$.

Similarly, you have $DB =DI$.

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Due to symmetry it suffices to show that $DI=DC$. Since $\overline{DI}$ and $\overline{DC}$ are two sides of the triangle $\Delta DCI$, it suffices to show that the triangle angles opposite these sides, namely $\angle DIC$ and $\angle ICD$, are congruent. Since $\angle DIC$ is an exterior angle of the triangle $\Delta AIC$, then, by the exterior angle theorem, $m\left(\angle DIC\right) = m\left(\angle ICA\right) + m\left(\angle IAC\right)$. Since $I$ is the incenter of triangle $\Delta ABC$, it is the intersection of this triangle's angle bisectors. Therefore $\angle ICA \cong \angle ICB$ and $\angle IAC \cong \angle BAD$. So we are left to show that $m\left(\angle DIC\right) = m\left(\angle ICB\right) + m\left(\angle BAD\right)$. But, since $\angle BAD$ and $\angle BCD$ are two inscribed angles of the circle through $A, B, C$ that are subtended on the same arc $\overset{\frown}{BD}$, they are congruent. Hence $m\left(\angle DIC\right) = m\left(\angle ICB\right) + m\left(\angle BCD\right) = m\left(\angle ICD\right)$.

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