Question

.
CABCD is cyclic quadrilateral, lines AB and DC intersect
in the point F and lines AD and BC interseer in the point
E Show that the circumcircles of AECF and ACDE
intersect in a point G on the line EF.​

Answer 1

Answer:

∠FBC=90° (Angle in the semi circle is 90°)Similarly, ∠FGC=90° ∴∠FBC+∠FGC=90°+90°=180°Hence, FBCG is a cyclic quadrilteral.Similarly, DEGC is a cyclic quadrilteral.∠FGC+∠EGC=180°−∠FBE+180°−∠EDF (FBGC and DEGC are cyclic quadrilateral)=∠ABE+∠ADF (Linear pair)=180° (ABCD is a cyclic quadrilateral)Hence, FGE is straight line.