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, $$BD+CE=BC\iff \frac{ac}{a+b}+\frac{ab}{a+c}=a\iff b^2+c^2-bc=a^2,$$ which is true. $\square$