Let $BC$ and $BD$ be the tangent lines to the circle with diameter $AC$. Let $E$ be the second point of intersection of line $CD$ and the circumscribed circle of triangle $ABC$. Prove that $CD= 2DE$. (Matthew Kurskyi)
Problem
Source: V.A. Yasinsky Geometry Olympiad 2023 IX p2 , Ukraine
Tags: geometry, equal segments
NO_SQUARES
14.12.2023 00:26
Let $F$ be a midpoint of $CD$. We need to prove that $DE=DF$ or, because $\angle ADF =90^{\circ}$, that $\angle AEC = \angle AFD$. But $\angle AEC = \angle ABC =90^{\circ}-\angle CAB =90^{\circ}-\angle DAF = \angle AFD$ because $AB$ is symedian in triangle $ACD$.
bin_sherlo
30.12.2023 20:47
Let $\angle CDB=\alpha$ and $\angle AEC=\beta$ and $AD\cap EB=M$ $\angle ADC=90$ so $\angle DAE=90-\beta=90-\angle ABC=\angle CEB\implies ME^2=MD.MA$ Also $\angle BAD=\alpha+\beta-90=\alpha-(90-\beta)=\angle EBD\implies MD.MA=MB^2$ $\implies ME=MB$ \[\frac{CD}{DB}=\frac{\sin 2\alpha}{\sin \alpha}=2\cos \alpha\]\[\sin \beta=\frac{DE}{EM}\]\[\frac{\cos \alpha}{\sin \beta}=\frac{BM}{BD}\]By multiplying these $3$ equalities, we get \[CD=2.DE\]as desired.
