Question

Quadrilateral ABCD is a cyclic quadrilateral lines ab and BC intersect in the. F and lines CD and BC intersect in the point if so that the circumcircle of triangle BCF and triangle CD intersect in a point P on the line EF

Answer 1

Proved below.

Step-by-step explanation:

Given:

Here quadrilateral ABCD is a cyclic quadrilateral.

As shown in the figure, AB and CD intersect in the point F and lines AD and BC intersect in the point E.

∠FBC = 90°        (Angle in the semi circle is 90°)

Similarly,

∠FPC = 90°  

∴ ∠FBC+∠FPC  = 90°+90° = 180°

Hence, FBCP  is a cyclic quadrilteral.

Similarly, DEPC is a cyclic quadrilteral.

∠FPC+∠EPC = 180°-∠FBE + 180°−∠EDF     (FBPC and DEPC are cyclic quadrilateral)

∠FPC+∠EPC = ∠ABE+∠ADF      (Linear pair)

∠FPC+∠EPC = 180°    (ABCD is a cyclic quadrilateral)

Hence, FPE is straight line.

Therefore, the circumcircle of triangle BCF and triangle CDE intersect in a point P on the line EF.

Hence proved.