Team Selection Test for JBMO 2023

Question

Let ABC be a triangle and D, E be points on segments AB, AC respectively, such that DE BC. Let the circumcircle of ABC meet the circumcircles of BDE and CDE again at K, L respectively. Let T be the intersection of the lines BK and CL. Prove that TA is tangent to the circumcircle of ABC.

Accepted Answer

The radical axes of the circles (ABC), (BDE), (CDE) must be concurrent at T; hence T, D, E are collinear. Moreover, TD TE = TB TK; hence the power of T with respect to the circles (ABC), (ADE) are equal and it lies on their radical axis. Since DE and BC are parallel, the radical axis of (ABC), (ADE) is the common tangent at A; hence TA is tangent to (ABC).