05/05/2026

$ABCD$ is a quadrilateral such that $\angle{ADB}=50^{\circ}$, $\angle{BDC}=30^{\circ}=\angle{ACB}$, $\angle{ACD}=40^{\circ}$, find $\angle{BAC}$

Solve:

\(\begin{multline} \shoveleft \text{Let }E \text{ be the circumcenter of }\triangle{BCD}, F=AC\cap BD\\ \shoveleft \implies EB=EC=ED, \angle{BEC}=2\angle{BCD}=60^{\circ}\\ \shoveleft \implies \triangle{BCE} \text{ is equilateral} \implies \\ \shoveleft \angle{ECF}=\angle{BCF}=30^{\circ}, \angle{DBE}=\angle{BDE}=20^{\circ} \implies \angle{ADE}=70^{\circ}\\ \shoveleft \implies \triangle{ABC}\cong\triangle{AEC},\triangle{BCF}\cong\triangle{ECF}\implies \angle{BAC}=\angle{EAC},BF=EF\\ \shoveleft \implies \angle{CFE}=180^{\circ}-30^{\circ}-60^{\circ}-20^{\circ}=70^{\circ}=\angle{ADE}\\ \shoveleft \implies ADEF \text{ is cyclic}\implies \angle{EAC}=\angle{BDE}=20^{\circ}\implies \angle{BAC}=\bbox[5px, border: 1px solid black]{20^{\circ}} \end{multline}\)

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