L OM - I - Task 3

In an isosceles triangle $ ABC $, angle $ BAC $ is a right angle. Point $ D $ lies on side $ BC $, such that $ BD = 2 \cdot CD $. Point $ E $ is the orthogonal projection of point $ B $ onto line $ AD $. Determine the measure of angle $ CED $.
Let's complete the triangle $ABC$ to a square $ABFC$. Assume that line $AD$ intersects side $CF$ at point $P$, and line $BE$ intersects side $AC$ at point $Q$. Since

 

 

$ CP= \frac{1}{2} CF $. Using the perpendicularity of lines $AP$ and $BQ$ and the above equality, we get $ CQ= \frac{1}{2} AC $, and consequently $ CP=CQ $. Points $C$, $Q$, $E$, $P$ lie on the same circle, from which $ \measuredangle CED =\measuredangle CEP =\measuredangle CQP = 45^\circ $.