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# Geometry 2

In $$\triangle ABC$$, $$\angle B >\angle C$$. Let $$P$$ and $$Q$$ be two points on line $$CA$$ such that $$\angle PBA=\angle QBA=\angle ACB$$ and $$A$$ is located between $$P$$ and $$C$$. Suppose that there exists an interior point $$D$$ of segment $$BQ$$ for which $$PD=DB$$. Let the ray $$AD$$ intersect the circumcircle of $$\triangle ABC$$ at $$R\neq A$$. Prove that $$QB=QR$$.

Note by Jubayer Nirjhor
3 years, 6 months ago

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