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