triangles
This questions is related to the question before. I was mistaken. Because I thought ED and AB are parallel, but they're not. So this question made me really confused and curious about it.
Given a Triangle ABC. AD, BE, and CF are altitudes. Prove that AD is the bisector of FDE angle.
It is easier to prove the following equivalent version:
Let $N$ be the point of intersection between the circumcircle of $ABC$ (centered at $O$) and the angle bisector from $A$. Then $N$ lies on the perpendicular bisector of $BC$.
Proof: since $\widehat{BAN}=\widehat{NAC}$, the $BN$ and $NC$ arcs in the circumcircle of $ABC$ have the same length, since they are subtended by equal angles at $O$. It follows that the segments $BN$ and $NC$ have the same length, and since $OB=OC=R$, the line through $O$ and $N$ is the perpendicular bisector of $BC$.
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
First, let's prove that triangle $CED$ is similar to $CBA$. Since $\angle AEB = 90^{\circ} = \angle ADB$, quadrilateral $ABDE$ is inscribed into a circle. Therefore, $\angle ABD = 180^{\circ} - \angle AED = \angle CED$. By analogy, $\angle BAE = \angle CDE$ what completes the proof. The same goes for similarity of $BDF$ and $BAC$. We want to prove that $\angle FDA = \angle ADE \Leftrightarrow 90^{\circ} - \angle FDA = 90^{\circ} - \angle ADE \Leftrightarrow \angle BDF = \angle CDE$. Since both of them are equal to $\angle BAC$ (due to lemma), proof is complete.