MathDB

4

Part of 2011 239 Open Mathematical Olympiad

Problems(2)

acute ABCD, BAD similar to BKL, then APK= LBC

Source: 239 2011 S4

5/19/2020
In convex quadrilateral ABCDABCD, where AB=ADAB=AD, on BDBD point KK is chosen. On KCKC point LL is such that △BAD∼△BKL\bigtriangleup BAD \sim \bigtriangleup BKL. Line parallel to DLDL and passes through KK, intersect CDCD at PP. Prove that ∠APK=∠LBC\angle APK = \angle LBC.
@below edited
geometry
Rombus ABCD, \angle AQO=\angle PBC

Source: 239 2011 J4

5/18/2020
Rombus ABCD with acute angle BB is given. OO is a circumcenter of ABCABC. Point PP lies on line OCOC beyond CC. PDPD intersect the line that goes through OO and parallel to ABAB at QQ. Prove that ∠AQO=∠PBC\angle AQO=\angle PBC.
geometry