Let $P$ be a point inside $\triangle{ABC}$ such that $\angle{ABP}=10^\circ, \angle{BAP}=\angle{BCP} = 20^\circ$. Show $\triangle{ABC}$ is an isosceles

Solution 1:

These conditions do not uniquely determine a triangle. By the inscibed angle theorem, $C$ could be anywhere on the green circle such that $P$ is still in $\triangle ABC$:

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Let $P$ be a point inside $\triangle{ABC}$ such that $\angle{ABP}=10^\circ, \angle{BAP}=\angle{BCP} = 20^\circ$. Show $\triangle{ABC}$ is an isosceles

Solution 1:

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