KöMaL 2026 B5508

Let $ABCD$ be a convex quadrilateral. a) Prove that if $\tan\angle BAC\cdot\tan\angle DCA =\tan\angle CAD\cdot\tan\angle ACB$, then $\tan\angle CBD\cdot\tan\angle ADB=\tan\angle DBA\cdot\tan\angle BDC$. b) Prove that if $\tan\frac{\angle BAC}{2}\cdot\tan\frac{DCA}{2}={\tan\frac{\angle CAD}{2}\cdot\tan\frac{\angle ACB}{2}}$, then ${\tan\frac{\angle CBD}{2}\cdot\tan\frac{\angle ADB}{2}}={\tan\frac{\angle DBA}{2}\cdot\tan\frac{\angle BDC}{2}}$.