Bulgarian National Mathematical Olympiad

Let $k=k(O,r)$ and let $k_1(I,r_1)$ touch $BD$, $CD$ and $k$ at $P$, $R$ and $Q$, respectively. Let $k_2(J,r_2)$ touch $AD$, $CD$ at $K$, $M$ and $L$, respectively. First, we shall prove that $AB\parallel IJ$, i.e. $ABIJ$ is an isosceles trapezoid. Assume the contrary and set $T=IJ\cap AB$. Then

$$\frac{IT}{TJ}\cdot \frac{JL}{LO}\cdot \frac{OQ}{QI}=\frac{r_1}{r_2}\cdot \frac{r_2}{r}\cdot \frac{r}{r_1}=1$$

and, by the Menelaus theorem, the points $T$, $Q$ and $L$ are collinear.



Then $TQ\cdot TL=TB\cdot TA$. On the other hand, $ABIJ$ is cocyclic which implies $TB\cdot TA=TI\cdot TJ$. It follows that $TQ\cdot TL=TI\cdot TJ$, i.e. $QLJI$ is cocyclic. But $\nparallel JLQ=\nparallel IQL$, i.e. $QLJI$ is an isosceles trapezoid and then $IJ\parallel QL$, a contradiction. Hence $AB\parallel IJ$, i.e. $r_1=r_2$ and $AK=BP$ (1).

Further, the generalized Ptolemy theorem (applied to $A$, $B$, $Q$ and $C$) gives $AB\cdot CR+AC\cdot BP=AP\cdot BC$. Since $CR=CD-DR=CD-DP=CD-BD+BP$ and $AP=AB-BP$, we get

$$(2)BP=\frac{AB(BC+BD-CD)}{AB+BC+AC}$$

Analogously,

$$(3)AK=\frac{AB(AC+AD-CD)}{AB+BC+AC}$$

Finally, (1), (2) and (3) imply $BC+BD=AC+AD$ which holds if and only if $D$ is the tangent point of $AB$ and the excircle to this side.