03/31/2026

$D$ is on side $BC$ of $\triangle{ABC}$ such that $\angle{BAD}=\angle{DAC}=20^{\circ}$, $\angle{ABC}=80^{\circ}$, $E$ is on $AD$ such that $BC=AE$, find $\angle{BCE}$

Solve 1:

\(\begin{multline} \shoveleft \angle{BAC}=40^{\circ},\angle{ABC}=80^{\circ}\implies \angle{BCA}=60^{\circ}\\ \shoveleft \text{Let }F \text{ on }AC \text{ such that }BC=BF \implies \triangle{ACF} \text{ is equilateral}\\ \shoveleft \implies BF=BC=AE, \angle{CBF}=60^{\circ}\implies \angle{ABF}=20^{\circ}=\angle{BAD}\\ \shoveleft \text{Let }AE \cap BF=G \implies AG=BG \implies EG=FG\\ \shoveleft \implies \angle{EFG}=\angle{FEG}=\angle{BAG}=\angle{ABG}=20^{\circ}\\ \shoveleft \implies ABEF \text{ is cyclic }\implies \angle{EBF}=\angle{EAF}=20^{\circ}\\ \shoveleft \implies BE \text{ bisects }\angle{ABC}\implies E \text{ is the incenter of }\triangle{ABC}\\ \shoveleft \implies \angle{BCE}=\dfrac{\angle{ACB}}{2}=\bbox[5px, border: 1px solid black]{30^{\circ}} \end{multline}\)

Solve 2:

\(\begin{multline} \shoveleft \text{Make equilateral }\triangle{ABF}\implies AB=BF, \angle{ABF}=\angle{BAF}=60^{\circ} \implies \\ \shoveleft \angle{CBF}=20^{\circ}=\angle{BAE}=\angle{CAF}\implies ABCF\text{ is cyclic}, \triangle{ABE}\cong\triangle{BCF}\\ \shoveleft \implies \angle{BFC}=\angle{ABE}=\angle{BAC}=40^{\circ}\implies E\text{ is incenter of }\triangle{ABC}\\ \shoveleft \implies \angle{BCE}=\dfrac{\angle{ACB}}{2}=\bbox[5px, border: 1px solid black]{30^{\circ}} \end{multline}\)

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