Question

Answer please
First answer will be brainlist.​

Answer 1

Answer:

ABCD is cyclic quadrilateral, line AB and DC intersect in the point F and lines AD and BC intersect in the point E. Show that the circumcircles of triangle BCF and triangle CDE intersect in a point G on the line EF. ​

Given,

1. ∠ FBC = 90°  (angle in a semi-circle is 90°)
2. Similarly, ∠ FGC = 90°

• ∴ ∠ FBC + ∠ FGC = 90° + 90° = 180°

Therefore, FBCG is a cyclic quadrilateral

Similarly,

∠ EDC = 90°  (angle in a semi-circle is 90°)

Similarly, ∠ EGC = 90°

∴ ∠ EDC + ∠ EGC = 90° + 90° = 180°

Therefore, DEGC is a cyclic quadrilateral

• From above equations, we have,

∠ FGC + ∠ EGC = 180° - ∠ FBE + 180° - ∠ EDF

= ∠ ABE + ∠ ADF (linear pair)

= 180°  (as ABCD is a cyclic quadrilateral)

• Hence, FGE is a straight line