06/09/2026

$P$ is a point inside $\triangle{ABC}$ such that $\angle{ABP}=40^{\circ}$, $\angle{PBC}=30^{\circ}$, $\angle{ACP}=10^{\circ}$, $\angle{BCP}=20^{\circ}$, find $\angle{BAP}$

Solve:

\(\begin{multline} \shoveleft \text{Easy to see }CP\perp AB, \angle{BAC}=80^{\circ}\\ \shoveleft \text{Make }BE \perp AC \text{ such that }\angle{ACE}=10^{\circ}, BE\cap CP=D, CP\cap AB=F\\ \shoveleft \implies D \text{ is the orthocenter of }\triangle{ABC} \implies \angle{BAD}=20^{\circ}, \angle{ABD}=10^{\circ} \\ \shoveleft\implies \angle{ADF}=70^{\circ},\angle{EBP}=30^{\circ}, \angle{DAC}=60^{\circ}\implies \angle{ADE}=30^{\circ}\\ \shoveleft AC\perp DE, \angle{ACE}=\angle{ACD}=10^{\circ}\implies \triangle{ACE}\cong\triangle{ACD}\\ \shoveleft \implies AD=AE \implies \angle{AED}=\angle{ADE}=30^{\circ}\\ \shoveleft \angle{EBP}=\angle{CBP}, \angle{ECP}=\angle{BCP}\implies P \text{ is the incenter of }\triangle{EBC}\\ \shoveleft \implies \angle{BEP}=\angle{CEP}=40^{\circ}\implies \angle{AEP}=70^{\circ}=\angle{ADF}\\ \shoveleft \implies ADPE \text{ is cyclic }\implies \angle{DAP}=\angle{BEP}=40^{\circ}\implies \angle{BAP}=\bbox[5px, border: 1px solid black]{60^{\circ}} \end{multline}\)