Saudi Arabian Mathematical Competitions

1. Denote by $G$, $H$ the intersections of $KE$, $KF$ and $BC$ respectively. Consider triangle $CEG$: it is easy to see that
- $CI$ is the internal bisector of $\angle C$
- $EI$ is the external bisector of $\angle E$.
Hence, $I$ is the excenter of this triangle, which implies that $GI$ is the bisector of $\angle KGH$.

Similarly, we can also see that $HI$ is the bisector of $\angle GHK$. So $I$ is the incenter of triangle $GHK$ and $KI$ is the bisector of $\angle GKH$.
In the other hand, $BA$, $BC$ are symmetric with respect to the line $BE$ and $CA$, $KG$ are also symmetric with respect to the line $BE$. From these facts, we can conclude that $\angle KGH = \angle BAC$.
Similarly, we also have $\angle KHG = \angle BAC$; therefore, $KGH$ is an isosceles triangle.
Since $I$ is the incenter of the isosceles triangle $KGH$, $KI$ is also an altitude, i.e. $KI \perp BC$.

2. Denote $BC = a$, $CA = b$, $AB = c$ and $A = \alpha$. Then $\angle DIE = 90^{\circ} + \frac{\alpha}{2}$.
In the isosceles triangle $KGH$, we have $\angle KGH = \angle KHG = \alpha$, so $\angle DKE = 180^{\circ} - 2\alpha$. Hence, if $I \in (KDE)$, then
$$90^{\circ} + \frac{\alpha}{2} + 180^{\circ} - 2\alpha = 180^{\circ}$$
and $\alpha = 60^{\circ}$. From the cosine law in triangle $ABC$, we have
$$a^{2} = b^{2} + c^{2} - bc .$$
By calculating, it is also easy to check that $BD = \frac{ac}{a+b}$, $CE = \frac{ab}{a+c}$; thus,
$$BD + CE = BC \Leftrightarrow \frac{ac}{a+b} + \frac{ab}{a+c} = a \Leftrightarrow b^{2} + c^{2} - bc = a^{2},$$
which is true. $\square$