2002 JBMO Problems/Problem 1
The triangle has . is a point on the circumcircle between and (and on the opposite side of the line to ). is the foot of the perpendicular from to . Show that .
Since is a cyclic quadrilateral, , and --- (1)
Now let be the foot of the perpendicular from to . Then we have, is a cyclic quadrilateral with as diameter of the circumcircle.
It follows that and are congruent (since ). So, we have and --- (2)
Also, in we have ( from (1) above) Thus So from , is congruent to . Hence we have .
Now, (from (2) above)