Baltic Way 2019

Let $D^{\prime }$ be the point symmetric to point $D$ with respect to line $AC$. As triangles $ADC$ and $A{D}^{\prime }C$ are similar, it suffices to prove that the inequalities $\angle CAB+\angle CA{D}^{\prime }<180^{\circ }$ and $\angle ACB+\angle AC{D}^{\prime }<180^{\circ }$ together imply $\angle ABC<\angle A{D}^{\prime }C$.

Suppose the contrary, i.e., $\angle A{D}^{\prime }C\le \angle ABC$. Then point $D^{\prime }$ is located on the circumcircle of triangle $ABC$ or outside it. Let $E$ be the second intersection point of line $A{D}^{\prime }$ with the circumcircle of triangle $ABC$ ($E=A$ if $A{D}^{\prime }$ is tangent). Similarly, let $F$ be the second intersection point of line $C{D}^{\prime }$ with the circumcircle of triangle $ABC$ ($F=C$ if $C{D}^{\prime }$ is tangent). Consider three cases that may occur.

If $\angle AC{D}^{\prime }\ge \angle ACB$ and $\angle CA{D}^{\prime }\ge \angle CAB$ then point $B$ is located inside triangle $A{D}^{\prime }C$ or on its side. Applying the triangle inequality twice now gives $|AB|+|BC|\le |A{D}^{\prime }|+|D^{\prime }C|$, whereas adding the two inequalities given in the problem results in $|AD|+|DC|<|AB|+|AC|$. This leads to contradiction as $|A{D}^{\prime }|=|AD|$ and $|D^{\prime }C|=|DC|$.

If $\angle AC{D}^{\prime }<\angle ACB$ then points $A,F,B,C$ occur on the circumcircle of triangle $ABC$ in this order whereas points $E$ and $B$ lie on different sides of line $C{D}^{\prime }$. Consequently, $$\angle FBC>\angle EBC=180^{\circ }-\angle D^{\prime }AC>\angle CAB=\angle CFB,$$ whence in triangle $FBC$ we get $|CF|>|CB|$ implying $|C{D}^{\prime }|>|CB|$, contradiction.

* The case $\angle CA{D}^{\prime }<\angle CAB$ is symmetric to the previous case.

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Alternative solution.

Since $AD<AB$, we have $\angle ABD<\angle ADB$. Similarly, $\angle CBD<\angle CDB$ because $CD<CB$. Hence $$\angle ABC=\angle ABD+\angle CBD<\angle ADB+\angle CDB=\angle ADC.$$