moldovan 23.07.2009 23:22 Let $ABCD$ be a convex pentagon such that \[\angle DCB = \angle DEA = 90^\circ, \ \text{and} \ DC=DE.\] Let $F$ be a point on AB such that $AF:BF=AE:BC$. Show that \[\angle FEC= \angle BDC, \ \text{and} \ \angle FCE= \angle ADE.\]
livetolove212 24.07.2009 04:02 Let $ (ADE)\cap EC=\{J\}. AJ\cap BC=\{I\}$ Then $ DJCI$ is cyclic quadrilateral we get $ \angle JID=\angle DCJ=\angle DEC=\angle DAJ$ So $ AD=DJ$ Thus $ \Delta DEA=\Delta DCI$ $ \Rightarrow CI=EA \Rightarrow \frac{BF}{FA}=\frac{BC}{CI}$ $ \Rightarrow FC//AI \Rightarrow \angle FCE=\angle CJI=\angle ADE$ Similarly, $ \angle FEC=\angle BDC$