circlesgeometrytriangles
If $I$ is the incentre of $\triangle ABC$ and $AI$ meets the circumcircle at $D$, the prove that $DB = DC = DI$.
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.
Let the midpoints of $CB$ and $CA$ be $A'$ and $B'$. Now we place a point on $D$ on $AB$ such that $AD=AB'$. And as $2c=a+b$, we also have $BD=BA'$.
Thus, by SAS congruence, $\triangle BIA'\cong \triangle BID$ and $\triangle AIB'\cong \triangle AID$. Thus, $\angle BDI= \angle IA'B$ and $\angle ADI =\angle IB'A$. Hence, $\angle IB'A=180^{\circ}-\angle IA'B=\angle IA'C$.
Thus, $C$, $B'$, $A'$ and $I$ lie on a circle.
Remark: The problem as stated is not entirely correct. Only when both $\angle ABC$ and $\angle ACB$ are acute angles do we have that $D$ is the incenter of the triangle $IJK$. When $\angle ABC$ or $\angle ACB$ is a right angle, we obtain a degenerate case, where $M=E=D=B$ so that $IJ=AB$, or $N=F=D=C$ so that $IJ=AC$, respectively. If $\angle ABC>\dfrac{\pi}{2}$ or $\angle ACB>\dfrac{\pi}{2}$, then $D$ is an excenter of the triangle $IJK$. See an illustration below. (If we define $I$ and $J$ so that $MI$ and $NJ$ are disjoint, then in the case $\angle ABC>\dfrac{\pi}{2}$, $D$ is the excenter of the triangle $IJK$ opposite to $I$. On the other hand, if $\angle ACB>\dfrac{\pi}{2}$, then $D$ is the excenter of the triangle $IJK$ opposite to $J$.)
Let $R$ denote the circumradius of the triangle $ABC$. Write $\alpha$, $\beta$, and $\gamma$ for the angles $\angle BAC$, $\angle CBA$, and $\angle ACB$, respectively. In what follows, we assume that $\beta$ and $\gamma$ are in the interval $\left(0,\dfrac{\pi}{2}\right)$. First, we assume that $\alpha\in\left(0,\dfrac{\pi}{2}\right)$. Then, $$\angle DFE=\pi-2\gamma\,.$$ Thus, the distance from $D$ to $EF$ is given by $$DF\,\sin(\pi-2\gamma)=DF\,\sin(2\gamma)\,.$$ Since the triangle $ABC$ and the triangle $DBF$ are similar, we get $$DF=AC\,\left(\frac{BD}{AB}\right)=AC\,\cos(\beta)=2\,R\,\sin(\beta)\,\cos(\beta)=R\,\sin(2\beta)\,.$$ Thus, the distance from $D$ to $EF$ is $R\,\sin(2\beta)\,\sin(2\gamma)$, whence the distance from $D$ to $IJ$ is $$\rho:=\frac{1}{2}\,R\,\sin(2\beta)\,\sin(2\gamma)\,.$$
It is not difficult to see that $OA\perp EF$, whence $OA \perp IJ$. Therefore, $A$ is the midpoint of the arc $IAJ$. This shows that $$\angle DKI=\angle AKI=\angle AKJ=\angle DKJ=:\theta\,.$$ Ergo, the distance between $O$ and $EF$ is $$\begin{align}d:=OA-AE\,\sin(\beta)&=R-AB\,\left(\frac{AE}{AB}\right)\,\sin(\beta)\\&=R-\big(2\,R\,\sin(\gamma)\big)\,\cos(\alpha)\,\sin(\beta)\\&=R\big(1-2\,\cos(\alpha)\,\sin(\beta)\,\sin(\gamma)\big)\,.\end{align}$$ Thus, the distance $\delta$ between $O$ and $IJ$ is $$\begin{align}\delta=d-\rho&=R\,\Big(1-2\,\cos(\alpha)\,\sin(\beta)\,\sin(\gamma)-2\,\cos(\beta)\,\cos(\gamma)\,\sin(\beta)\,\sin(\gamma)\Big) \\ &=R\,\Big(1+2\,\big(\cos(\beta+\gamma)-\cos(\beta)\,\cos(\gamma)\big)\,\sin(\beta)\,\sin(\gamma)\Big) \\ &=R\,\Big(1-2\,\sin^2(\beta)\,\sin^2(\gamma)\Big)\,.\end{align}$$
Finally, note that $$\begin{align}DK=DH&=BD\,\cot(\gamma)=\big(AB\,\cos(\beta)\big)\,\cot(\gamma)\\&=\big(2\,R\,\sin(\gamma)\big)\,\cos(\beta)\,\cot(\gamma)\\&=2\,R\,\cos(\beta)\,\cos(\gamma)\,.\end{align}$$ Since $\delta=R\,\cos(2\,\theta)=R\,\big(1-2\,\sin^2(\theta)\big)$, we conclude that $$\sin(\theta)=\sin(\beta)\,\sin(\gamma)\,,$$ whence the distance from $D$ to $IK$ or to $JK$ is $$DK\,\sin(\theta)=DK\,\sin(\beta)\,\sin(\gamma)=2\,R\,\cos(\beta)\,\cos(\gamma)\,\sin(\beta)\,\sin(\gamma)=\rho\,.$$ Hence, $D$ is the incenter of the triangle $IJK$, with inradius $\rho=\dfrac{1}{2}\,R\,\sin(2\beta)\,\sin(2\gamma)$.
The case $\alpha \in\left(\dfrac{\pi}{2},\pi\right)$ can be dealt with in a similar manner. One of the differences is that $\angle DFE=2\gamma$.
Best Answer
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 $DB =DI$.