Thailand Mathematical Olympiad

We use Kelly's lemma

Lemma 1 (Kelly). Given a triangle $ABC$. Suppose the cevians $BE$ and $CF$ are such that $\angle CBE\ge \angle BCF$ and $\angle ABE\ge \angle ACF$. Then $BE\le CF$.

Proof (From Crux). Choose $Q$ on the segment $AE$ so that $\angle QBE=\angle QCF$. Let $CF$ meets $BE,BQ$ at $P,Q$ respectively. In the triangle $QBC$, since $\angle QBC\ge \angle QCB$, we have $QC\ge QB$. Observe that $△QBE\sim △QCR$, hence from that $BQ\le CQ$ we obtain $BE\le CR$. Clearly $CR\le CF$, therefore $BE\le CF$. $\square$

Now we apply Kelly's lemma to our problem. We want to show that $$\angle DBC\ge \angle ECB\text{and}\angle ABP\ge \angle ACP.$$

Reflect the point $C$ about the line $AP$ to $C^{\prime }$. By symmetry $$\angle ABP\ge \angle A{C}^{\prime }P=\angle ACP.$$ To show that $\angle DBC\ge \angle ECB$, we use sine law in the triangles $ABP$ and $ACP$ respectively to get $$BP=AP\frac{\sin (A/2)}{\sin (\angle ABP)},CP=AP\frac{\sin (A/2)}{\sin (\angle ACP)}.$$ Thus $BP\le CP$. In the triangle $PBC$, since $BP\le CP$, it follows that $\angle PCB\le \angle PBC$. Therefore $$\angle DBC\ge \angle ECB.$$ This proves the claim and the problem.