Let $P$ be a point on circumcircle of triangle $ABC$ on arc $\stackrel{\frown}{BC}$ which does not contain point $A$. Let lines $AB$ and $CP$ intersect at point $E$, and lines $AC$ and $BP$ intersect at $F$. If perpendicular bisector of side $AB$ intersects $AC$ in point $K$, and perpendicular bisector of side $AC$ intersects side $AB$ in point $J$, prove that: ${\left(\frac{CE}{BF}\right)}^2=\frac{AJ\cdot JE}{AK \cdot KF}$
Problem
Source: Regional Olympiad - Federation of Bosnia and Herzegovina 2018
Tags: geometry, circumcircle, perpendicular bisector
Tintarn
19.09.2018 14:44
This is very easy using trigonometry. Let $\alpha, \beta, \gamma$ be the angles at $A, B, C$ in the cyclic quadrilateral $ABCP$ (so that $\beta+\gamma=180^\circ$) and let $b=AC, c=AB$. By the sine law, $\frac{CE}{b}=\frac{\sin \alpha}{\sin \alpha+\gamma}$ and $\frac{BF}{c}=\frac{\sin \alpha}{\sin \alpha+\beta}$. Moreover, by definition we have that $AK=\frac{c}{2\cos \alpha}$ and $AJ=\frac{b}{2\cos \alpha}$ as well as $\frac{AE}{b}=\frac{\sin \gamma}{\sin \alpha+\gamma}$ and $\frac{AF}{c}=\frac{\sin \beta}{\sin \alpha+\beta}$, again from the sine law.
After clearing $\frac{b^2}{c^2}$ from our equation we are then left to prove that
\[\frac{\sin (\alpha+\beta)^2}{\sin (\alpha+\gamma)^2}=\frac{\frac{1}{2\cos \alpha}-\frac{\sin \gamma}{\sin \alpha+\gamma}}{\frac{1}{2\cos \alpha}-\frac{\sin \beta}{\sin \alpha+\beta}}\]Multiplying both sides with $\frac{\sin \alpha+\gamma}{\sin \alpha+\beta}$ we need to prove that
\[\frac{\sin \alpha+\beta}{\sin \alpha+\gamma}=\frac{\sin (\alpha+\gamma)-2\cos \alpha\sin \gamma}{\sin(\alpha+\beta)-2\cos \alpha\sin\beta}\]But $\sin (\alpha+\gamma)-2\cos\alpha\sin \gamma=\sin \alpha\cos\gamma-\cos\alpha\sin \gamma=\sin(\alpha-\gamma)=\sin(\alpha+\beta)$ and similarly $\sin (\alpha+\beta)-2\cos\alpha\sin \beta=\sin (\alpha-\beta)=\sin (\alpha+\gamma)$ and we are done.