I need these answers urgently.

I need these answers urgently. 5. Given: APIBC and BOX AC. CF is a diameter of the circle. a) ABDE and GDCE are cyclic Prove: quads. b) PC=CQ. c) AFBG is a parallelogram. c

Dear student,

In quadrilateral GDCE we have ∠ GDC = 90^{\circ}, ∠ GEC = 90^{\circ} ∠ GDC + ∠ GEC = 180^{\circ} Hence GDCE is cyclic quadrilateral ∠ 1 + ∠ 2 + ∠ 3 = 90^{\circ} Also ∠ 2 + ∠ 3 + ∠ 4 = 90^{\circ} ∠ 1 + ∠ 2 + ∠ 3 = ∠ 2 + ∠ 3 + ∠ 4 i . e ∠ 1 = ∠ 4 Hence ABDE is cyclic quadrilateral Also we know that mirror image of orthocentre in side of triangle lies on circumcircle So GE = EQ and GD = DP Now in ΔCPD and ΔCGD we have GD = DP CD = CD ∠ CDP = ∠ CDG So ΔCPD ≅ ΔCGD and Hence CP = CG Similarly QC = CG Hence PC = QC Also we have ∠ FAC = 90^{\circ} Angle in a semicircle ∠ FAB + ∠ 3 + ∠ 4 = 90 So ∠ FAB = ∠ 2 Hence FA | | BG Similarly BF | | AD and Hence AFBG is a parallelogram Regards