Final Round of National Olympiad

a.
Consider the triangles $ADB$ and $CDB$ (Fig. 21). The claim $AD < AB$ implies $\angle ABD < \angle ADB$ because the longer side is opposite to the larger angle. Similarly, $CD < CB$ implies $\angle CBD < \angle CDB$. As $ABCD$ is convex, $\angle ABD + \angle CBD = \angle ABC$ and $\angle ADB + \angle CDB = \angle ADC$. Hence, adding the two inequalities gives $\angle ABC < \angle ADC$.

Fig. 21

b.
Let points $A, B, C$ be such that $AB = BC > AC$. Choose point $D'$ on the line tangent to the circumcircle of the triangle $ABC$ at $A$ in such a way that $B$ and $D'$ lie on the same side of $AC$ and the inequalities $AD' < AB$ and $CD' < BC$ hold (the last inequality is possible since $AC < BC$); let $D$ be the reflection of $D'$ from $AC$ (Fig. 22). Then both assumptions $AD < AB$ and $CD < CB$ hold. But the claim $\angle ABC < \angle ADC$ is not true: since $D'$ lies outside the circumference of triangle $ABC$, we have $\angle ABC > \angle AD'C = \angle ADC$. Hence the hypothesis does not hold for non-convex quadrilaterals.

Fig. 22