RMO 2018 Tamil Nadu Problem 1 is from Geometry. We present sequential hints for this problem. Do not read all hints at one go. Try it yourself.
Let ABC be an acute-angled triangle and let D be an interior point of the line segment BC. Let the circumcircle of triangle ACD intersect AB at E (E between A and B) and let the circumcircle of triangle ABD intersect AC at F (F between A and C). Let O be the circumcenter triangle AEF. Prove that OD bisects ( \angle EDF )
Join OE and OF. What kind of quadrilateral is OEDF? (Pause here. Try it yourself.)
Note that $ EDCA$ is a cyclic quadrilateral (all of its four vertices are on a circle). Hence $ \angle ODB = \angle EAC = \angle A $ .
Similarly $AFDB $ is cyclic. Hence $\angle CDF = \angle FAB = \angle A $
This implies $ \angle EDF = 180^o - 2 \angle A $. Since O is the center of $ \Delta AEF $, $ \angle EOF = 2\times \angle EAF = 2 \times \angle A$.
Therefore $ \angle EDF + \angle EAF = 180^o - 2 \times \angle A + 2 \times \angle A = 180^o $. This implies quadrilateral OEDF is cyclic.
OE = OF because O is the center and E, F are at the circumference of the circle passing through AEF. Therefore both are radii of the same circle.
This implies $ \angle OEF = \angle OFE $
Since OEDF is cyclic, hence $\angle OEF = \angle ODF $ (angle subtended by the arc OF at the circumference).
Similarly $ \angle OFE = \angle ODE $.
But $ \angle OEF = \angle OFE $ (see hint 4).
Hence $ \angle ODE = \angle ODF $. The proof is complete,