MathSolver94 wrote: Let $ABC$ be a triangle in which $AB=AC$ and let $I$ be its incenter. It is known that $BC=AB+AI$. Let $D$ be a point on line $BA$ extended beyond $A$ such that $AD=AI$. Prove that $DAIC$ is a cyclic quadrilateral. As before $AIC=90+0.5B$ and $BDC=90-0.5B$

$ABC$ and $BDC$ are isosceles implies that $I$ is the circumcentre of $BDC$ and $IB=ID\Longrightarrow \angle ADI=\angle ABI=\angle IBC=\angle ICB=\angle ACI$ and so, $ADCI$ is isosceles. Note: You can prove that $\angle BAC=\frac{\pi}{2}$.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1695934&sid=16e640c7be688190ca11d147e1d9a550#p1695934

Let $\hat B=\hat C=\alpha$. By symmetry, $BI_CI_DI$ and $I$ is the circumcenter of $\Delta BCD$, consequently $\angle DIC=2\alpha=\angle DAC$ and $DAIC$ is cyclic. Further, making a connection with the link above: from $\Delta DAI$ isosceles and with $\angle IDA=\angle ICA=\angle ADI$we get $\angle BAI=\hat C$; with $AI\bot BC$ we get $2\hat B=90^\circ\iff\hat A=90^\circ$. Best regards, sunken rock

Let $\angle BDC=\angle BCD=\alpha ;\angle ABC=\angle ACB=\beta$ By angles: $2\alpha+\beta=180 \implies \alpha+\frac{\beta}{2}=90$ Now: $\implies \angle ICD=\alpha-\frac{\beta}{2}$ $\implies \angle IAB=90-\beta$ $\alpha-\frac{\beta}{2}=90-\beta$ Thus: $DAIC$ is a cyclic quadrilateral.$\blacksquare$

BC=AB+AI=AB+AD=BD Let BC=BD=x° angle CAD=angle B+ angle C since, ∆ ABC is isosceles, AB= AC, angle B= angle C angle CAD=2angle In ∆ADC, angle DCA=180°-(2B+x) =A-x[ In ∆ ABC, 2B=180°-A] angle ACI= B/2 angle DCI= B/2+A-x angle CAI=A/2 angle DAI= angle DAC+ angle CAI =2B+A/2 In quadrilateral , ADCI angle DAI+ angle DCI=2B+(A/2)+(B/2)+A-x =(5B+3A)/2-x=[5B+3(180°-2B)]/2-x = 270°-(B/2+x) =270°-90°[In ∆ BDC, 2x+B=180°,x+B/2=90°] =180° So, we can conclude, ADCI is cyclic