51st Ukrainian National Mathematical Olympiad, 3rd Round

a) Using the properties of inscribed angles we get $KM\perp BD$, $KN\perp AC$, $\angle ABD=\angle ACD\Rightarrow \angle AMK=\angle ABD=\angle ACD=\angle KND$. Thus $\angle KMN=\pi -\angle AMK-\angle BMN=\pi -\angle KND-\angle MNC=\angle KNM$, hence $\square KMN$ is isosceles and $KM=KN$, which implies that $KMLN$ is rhombus and $\angle NKL=\angle NLK$ (fig. 8). Since by analogy we have that $\angle AMK=\angle KND$, $\angle DKN=\angle NLC$, then $\angle KLD=\angle DKN+\angle NKL=\angle NLC+\angle LNK=\angle CLK$, which proves part a).

b) Since $KM=KN$, then $DB=2KM=2KN=AC$ and $\angle ABC+\angle DAB=\pi$. We have $DA\perp BC$, or $\angle ABC=\angle DAB$. But since $\angle DAC=\angle DBC$, then $\angle CAB=\angle DAB-\angle DAC=\angle ABC-\angle DBC=\angle ABD=\angle ACD$, and $AB\perp BC$.