Problem 1694

We use Kelly's lemma

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

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 $\angle QBC \geq \angle QCB$, we have $QC \geq QB$. Observe that $\triangle QBE \sim \triangle QCR$, hence from that $BQ \leq CQ$ we obtain $BE \leq CR$. Clearly $CR \leq CF$, therefore $BE \leq CF$. $\square$

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

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